NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations: Overview, Questions, Preparation

Exercise 7.1

Q1. How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that repetition of the digits is allowed?

A.1. i.Since repetition of number is allowed and there are five numbers which can be used to form the necessary 3-digit numbers we can have five numbers that can fill the ones, tens and hundreds place.

So, total number of possible 3-digit number = 5 × 5 × 5 = 125

ii.Since repetition of number is not allowed. There are total 5 numbers which can fill the ones places then 4 and 3 numbers which can fill the tens and hundreds place simultaneously.

So, total number of possible 3-digit number = 3 × 4 × 5 = 60

 

Q2. How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?

A.2. Since we are given with six numbers (1, 2, 3, 4, 5, 6) to form 3-digit even number and also repetition is allowed.

We can have only the number 2, 4, 6 in the ones place and all the six numbers can fill the tens and hundreds place.

So, total number of 3-digit even number that can be formed by 1, 2, 3, 4, 5, 6

= 6 × 6 × 3

= 108

 

Q3. How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?

A.3. To form a 4-letter code using the first 10 letters of the English alphabet without repeating we can have 10, 9, 8 and 7 numbers of letters to be filled at ones, tens, hundreds and thousands place simultaneously.

Hence, total no. of 4-letter code that can be made using the first 10-letter of English alphabet = 7 × 8 × 9 × 10 = 5040

 

Q4. How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?

A.4. For the 5-digit telephone number that can be constructed using 0 to 9 if each number starts with 67 and no digit appears more than once we can have

6

7

6 numbers

7 numbers

8 numbers

So total number of possible combination = 1 × 1 × 6 × 7 × 8 = 336

 

Q5. A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?

A.5. When a coin is tossed one time a head or a tail is the possible outcome.

So, when a coin is tossed 3 times the total number of possible outcomes

= 2 × 2 × 2 = 8

 

Q6. Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?

A.6. To generate a signal which requires 2 flags one below another we can have the following combination of any of the 5 flag at top and the one of the remaining 4 flag at the bottom.

Hence, total no. of possible combination = 5 × 4 = 20

Exercise 7.2

Q1. Evaluate

(i) 8 ! (ii)4 ! – 3 !

A.1. We know that, n! = n(n – 1)(n – 2)……..

i. 8!

8! = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8

= 40320

ii. 4! – 3!

= (1 × 2 × 3 × 4) – (1 × 2 × 3)

= 24 – 6

= 18

 

Q2. Is 3 ! + 4 ! = 7 ! ?

A.2. L.H.S = 3! + 4!

= (1 × 2 × 3) + (1 × 2 × 3 × 4)

= 6 + 24

= 30

R.H.S = 7!

= 1 × 2 × 3 × 4 × 5 × 6 × 7

= 5040

As, L.H.S ≠ R.H.S

3! + 4! ≠ 7!

 

Exercise 7.3

Q1. How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit isrepeated?

A.1. The permutation of 9 different digits taken 4 at a time is given by

⁹P₃ = $\frac{9!}{\left(9-3\right)!}$ =  $\frac{9!}{6!}$ =  $\frac{\mathrm{9\prime }<br>\mathrm{8‘}<br>\mathrm{7’}<br>6!}{6!}$ = 9 × 8 × 7 = 504

 

Q2. How many 4-digit numbers are there with no digit repeated?

A.2. For every four-digit number we have to count the permutation of 10 digits namely 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 taken 4 at a time

i.e. ¹⁰P₄ = $\frac{10!}{\left(10-4\right)!}$ =  $\frac{1\mathrm{0‘} <br>\mathrm{9‘}<br>\mathrm{8\prime }<br>\mathrm{7‘}<br>6!}{6!}$ = 10 × 9 × 8 × 7 = 5040

However, these permutation will include those where 0 is at 1000’s place.

So, fixing 0 at 1000’s place and rearranging the remaining 9 digits taking 3 at a time.

i.e. ⁹P₃= $\frac{9!}{\left(9-3\right)!}$ =  $\frac{9!}{6!}$ =  $\frac{\mathrm{9\prime }<br>\mathrm{8\prime }<br>\mathrm{7\prime }<br>6!}{6!}$ = 9 × 8 × 7 = 504

Therefore, The required number = ¹⁰P₄ – ⁹P₃

= 5040 – 504

= 4536 ways

 

Q3. How many 3-digit even numbers can be made using the digits1, 2, 3, 4, 6, 7, if no digit is repeated?

A.3. The permutation of having even number at the last digit from the given 6 different digits namely 1, 2, 3, 4, 5, 6 to form a 3-digit number is

³P₁= $\frac{3!}{\left(3-1\right)!}$ =  $\frac{3!}{2!}$ =  $\frac{\mathrm{3\prime } <br>2!}{2!}$ = 3

After taking one of the even number as last digit we can rearrange the remaining 5 digits taking 2 at a time. i.e.

⁵P₂ = $\frac{5!}{\left(5-2\right)!}$ =  $\frac{5!}{3!}$ =  $\frac{\mathrm{5‘} <br>\mathrm{4\prime }<br>3!}{3!}$ = 5 × 4 = 20

Therefore, The required number = 20 × 3 = 60

Q4. Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4,5 if no digit is repeated. How many of these will be even?

A.4. The permutation of 5 different digits namely 1, 2, 3, 4, 5 taken 4 at a time is

⁵P₄ = $\frac{5!}{\left(5-4\right)!}$ =  $\frac{5!}{1!}$ = 5 × 4 × 3 × 2 × 1 = 120

The permutation of having 2 or 4 at ones place is

²P₁ = $\frac{2!}{\left(2-1\right)!}$ =  $\frac{2!}{1!}$ = 1 × 2 = 2

After fixing one of the even number at last digit we can rearrange the remaining four digits taking 3 at a time. i.e.

⁴P₃ = $\frac{4!}{\left(4-3\right)!}$ =  $\frac{4!}{1!}$ = 4 × 3 × 2 × 1 = 24

Therefore, total permutation of 4 digit even number using 1, 2, 3, 4, 5

= 24 × 2

= 48

 

Q5.From a committee of 8 persons, in how many ways can we choose a chairmanand a vice chairman assuming one person can not hold more than one position?

A.5.The permutation of 8 persons taken 2 positions at a time is

⁸P₂ = $\frac{8!}{\left(8-2\right)!}$ =  $\frac{\mathrm{8‘}<br>\mathrm{7\prime }<br>6!}{6!}$ = 8 × 7 = 56

 

 

 

=>  $\frac{5!}{\left(5-r\right)!}$ =  $\frac{6!}{\left(6-\left(r-1\right)\right)!}$

=>  $\frac{\left(6-r+1\right)!}{\left(5-r\right)!}$ =  $\frac{6!}{5!}$

=>  $\frac{\left(7-r\right)!}{\left(5-r\right)!}$ =  $\frac{6\times 5!}{5!}$

=>  $\frac{\left(7-r\right)\left(6-r\right)\left(5-r\right)!}{\left(5-r\right)!}$ = 6

=> (7 – r)(6 – r) = 6

=> 42 – 13r + r² – 6 = 0

=>r² – 13r + 36 = 0

=>r² – 9r – 4r + 36 = 0

=>r(r – 9) – 4(r – 9) = 0

=> (r – 4)(r – 9) = 0

=>r = 4, 9

From, , ⁵P_r, r ≤ 5

We have, r = 4

 

Q8. How many words, with or without meaning, can be formed using all the letters ofthe word EQUATION, using each letter exactly once?

A.8. Since no letter is repeated in the word EQUATION.

The permutation of 8 letters taken all at a time

= ⁸P₈

=  $\frac{8!}{\left(8-8\right)!}$

=  $\frac{8!}{0!}$

= 8! [since, 0! = 1]

= 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1

= 40320

 

Q9. How many words, with or without meaning can be made from the letters of theword MONDAY, assuming that no letter is repeated, if.

4 letters are used at a time,

all letters are used at a time,

all letters are used but first letter is a vowel?

A.9. i.The permutation of 6 letters in MONDAY taken 4 at a time without repetition is

⁶P₄ = $\frac{6!}{\left(6-4\right)!}$ =  $\frac{\mathrm{6‘}<br>\mathrm{5\prime }<br>\mathrm{4\prime }<br>\mathrm{3\prime }\times 2!}{2!}$ = 360

ii.The permutation of 6 letters in MONDAY when all letters are taken at a time is

⁶P₆ = $\frac{6!}{\left(6-6\right)!}$ =  $\frac{6!}{0!}$ = 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

iii.The permutation of having one of the two vowels (O, A) as first letter from the word MONDAY when all letters are taken at a time is

²P₁ = $\frac{2!}{\left(2-1\right)!}$ = 2

After fixing one of the vowel as first letter we can rearrange the remaining 5 letters taking 5 at a time

⁵P₅ = $\frac{5!}{\left(5-5\right)!}$ =  $\frac{5!}{0!}$ = 5! = 5 × 4 × 3 × 2 × 1 = 120

Therefore, total permutation when all letters are used but first letter is vowel from the word MONDAY = 2 × 120 = 240

 

Q10. In how many of the distinct permutations of the letters in MISSISSIPPI do thefour I’s not come together?

A.10. There are 11 letters of which M appears 1 time, I appears 4 times, S appears 4 times and P appears 2 times.

The required number of arrangements =  $\frac{11!}{4!4!2!}$

=  $\frac{1\mathrm{1\prime }<br>1\mathrm{0\prime }<br>\mathrm{9\prime }<br>\mathrm{8\prime }<br>\mathrm{7\prime }<br>\mathrm{6\prime }<br>\mathrm{5\prime }<br>4!}{\mathrm{4\prime }<br>\mathrm{3\prime }<br>\mathrm{2\prime }<br>\mathrm{1\prime }<br>4!\times 2}$

= 11 × 10 × 9 × 5

= 34650

When the four I occurs together we treat them as single object IIII. This single object together with 7 remaining object will account for 8 object which have 1-M. 2-P and 4-S.

So, required number of permutation =  $\frac{8!}{4!2!}$

=  $\frac{\mathrm{8\prime }<br>\mathrm{7\prime }<br>\mathrm{6\prime }<br>\mathrm{5\prime }<br>4!}{4!\prime 2}$

= 840

Therefore, total no. of permutation in which 4-I’s do not come together

= 34650 – 840

= 33810

 

Q11. In how many ways can the letters of the word PERMUTATIONS be arranged if the

words start with P and end with S,

vowels are all together,

thereare always 4 letters between P and S?

A.11. There are 12 letters in which T appears 2 times and rest are all different.

i.When P and S are fixed as first and last letter we can arrange the remaining 10 letter taking all at a time. i.e.

Number of permutation =  $\frac{10!}{2!}$

=  $\frac{1\mathrm{0\prime }<br>\mathrm{9\prime }<br>\mathrm{8\prime }<br>\mathrm{7\prime }<br>\mathrm{6\prime }<br>\mathrm{5\prime }<br>\mathrm{4\prime }<br>\mathrm{3\prime }<br>2!}{2!}$

= 18,14,400

ii.We take the 5 vowels (E,U,A,I,O) as one single object. This single object with the remaining 7 object are treated as 8 object which have 2 – T’s.

So, number of permutations in which the vowels come together

= permutation of 8 object x permutation within the vowels

=  $\frac{8!}{2!}$ × 5!

=  $\frac{\mathrm{8\prime } <br>\mathrm{7\prime }<br>\mathrm{6\prime }<br>\mathrm{5\prime }<br>\mathrm{4\prime }<br>\mathrm{3\prime }<br>2!}{2!}$ × 5 × 4 × 3 × 2 × 1

= 20160 × 120

= 2419200

iii.In order to have 4 letters between P and S, (P, S) should have the possible sets of places (1, 6), (2, 7), (3, 8), (4, 9), (5, 10), (6, 11) and (7, 12) similarly (S, P) should have the same possible sets of places.

So, total number of possible positions fixed for P and S = 14

After fixing P and S we can rearrange the remaining 10 places with the remaining 10 letters of which 2 are T’s.

So, required permutation =  $\frac{10!}{2!}$

=  $\frac{1\mathrm{0\prime }<br>\mathrm{9\prime }<br>\mathrm{8\prime }<br>\mathrm{7\prime }<br>\mathrm{6\prime }<br>\mathrm{5\prime }<br>\mathrm{4\prime }<br>\mathrm{3\prime }<br>2!}{2!}$

= 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3

= 18,14,400

Therefore, total permutation for which there are always 4 letters between P and S

= 14 × 18,14,400

= 2,54,01,600

 

Exercise 7.4

Q1. ⁿC₈ = ⁿC₂, find ⁿC2.

A.1. ⁿC₈ = ⁿC₂

As, ⁿC_{a =} ⁿC_b

=>a = b or a = n – b

=>n = a + b

We have,

ⁿC₈ = ⁿC₂

=>n = 8 + 2

=>n = 10

Therefore,

ⁿC₂

= ⁿC₂

=  $\frac{10!}{2!\left(10-2\right)!}$

=  $\frac{10!}{2!8!}$

=  $\frac{1\mathrm{0\prime }<br>\mathrm{9\prime }<br>8!}{2!8!}$

=  $\frac{1\mathrm{0\prime }<br>9}{2}$

= 45

 

Q2. Determine n if

²ⁿC₃ : ⁿC₃ = 12 : 1(ii)²ⁿC₃ : ⁿC₃ = 11 : 1

A.2. i.²ⁿC₃ : ⁿC₃ = 12 : 1

=>  $\frac{2n!}{3!\left(2n-3\right)!}$ ÷  $\frac{n!}{3!\left(n-3\right)!}$ =  $\frac{12}{1}$

=>  $\frac{2n\left(2n-1\right)\left(2n-2\right)\left(2n-3\right)!}{3!\left(2n-3\right)!}$ ×  $\frac{3!\left(n-3\right)!}{n\left(n-1\right)\left(n-2\right)\left(n-3\right)!}$ = 12

=>  $\frac{2n\left(2n-1\right)\left(2n-2\right)}{n\left(n-1\right)\left(n-2\right)}$ = 12

=>  $\frac{2\left(2n-1\right)2\left(n-1\right)}{\left(n-1\right)\left(n-2\right)}$ = 12

=> 4(2n - 1) = 12(n – 2)

=> 8n – 4 = 12n – 24

=> 24 – 4 = 12n – 8n

=> 20 = 4n

=>n =  $\frac{20}{4}$

=>n = 5

ii.²ⁿC₃ : ⁿC₃ = 11 : 1

=>  $\frac{2n!}{3!\left(2n-3\right)!}$ ÷  $\frac{n!}{3!\left(n-3\right)!}$ =  $\frac{11}{1}$

=>  $\frac{2n\left(2n-1\right)\left(2n-2\right)\left(2n-3\right)!}{3!\left(2n-3\right)!}$ ×  $\frac{3!\left(n-3\right)!}{n\left(n-1\right)\left(n-2\right)\left(n-3\right)!}$ = 11

=>  $\frac{2n\left(2n-1\right)2\left(n-1\right)}{n\left(n-1\right)\left(n-2\right)}$ = 11

=> 4(2n – 1) = 11(n – 2)

=> 8n – 4 = 11n – 22

=> 22 – 4 = 11n – 8n

=> 18 = 3n

=>n =  $\frac{18}{3}$

=>n = 6

 

Q3. How many chords can be drawn through 21 points on a circle?

A.3. A chord is drawn by connecting 2 points on a circle.

As we are given with 21 points on the circle, we have the following combination to find the number of chords.

²¹C₂ = $\frac{21!}{2!\left(21-2\right)!}$ =  $\frac{2\mathrm{1\prime }<br>2\mathrm{0\prime }<br>19!}{\mathrm{2\prime }<br>\mathrm{1\prime }<br>19!}$ = 210

 

Q4. In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and4 girls?

A.4.The number of ways of selecting a team consisting of 3 boys from 5 boys and 3 girls from 4 girls is

⁵C₃×⁴C₃

=  $\frac{5!}{3!\left(5-3\right)!}$ ×  $\frac{4!}{3!\left(4-3\right)!}$

=  $\frac{\mathrm{5\prime }<br>\mathrm{4\prime }<br>3!}{3!\prime <br>2!}$ ×  $\frac{\mathrm{4\prime }<br>3!}{3!\prime <br>1!}$

=  $\frac{20}{2}$ ×  $\frac{4}{1}$

= 40

 

Q5. Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5blue balls if each selection consists of 3 balls of each colour.

A.5. Since we are to select 3 balls from each colour in order to select 9 balls from the collections of 6, 5 and 5 balls of red, white and blue colours respectively, we can have the combination

⁶C₃(red) ×⁵C₃ (white) ×⁵C₃ (blue)

=  $\frac{6!}{3!\left(6-3\right)!}$ ×  $\frac{5!}{3!\left(5-3\right)!}$ ×  $\frac{5!}{3!\left(5-3\right)!}$

=  $\frac{\mathrm{6\prime }<br>\mathrm{5\prime }4\prime <br>3!}{\mathrm{3\prime }<br>\mathrm{2\prime }<br>3!}$ ×  $\frac{\mathrm{5\prime }<br>\mathrm{4\prime }<br>3!}{3!<br>2!}$ ×  $\frac{\mathrm{5\prime }<br>\mathrm{4\prime }<br>3!}{3!<br>2!}$

= 20 × 10 × 10

= 2000

 

Q6. Determine the number of 5 card combinations out of a deck of 52 cards if thereis exactly one ace in each combination.

A.6. In a deck of 52 cards there are 4 ace cards. The required number of ways of selecting one ace card from the four = ⁴C₁ = $\frac{4!}{1!\left(4-1\right)!}$ =  $\frac{4!}{3!}$ =  $\frac{4\times 3!}{3!}$ = 4

After selecting one ace we need to select the remaining 4 card from the remaining 48 card to have a combination of 5 cards. The required number of ways

= ⁴⁸C₄

=  $\frac{48!}{4!\left(48-4\right)!}$

=  $\frac{4\mathrm{8\prime }<br>4\mathrm{7\prime }<br>4\mathrm{6\prime }<br>4\mathrm{5\prime }<br>44!}{4!\prime <br>44!}$

=  $\frac{4\mathrm{8\prime }<br>4\mathrm{7\prime }<br>4\mathrm{6\prime }<br>45}{\mathrm{4\prime }<br>\mathrm{3\prime }<br>2}$

= 1,94,580

Therefore, the total number of ways for selecting 5 card combination out of a deck of 52 cards if there is exactly one ace in each combination

= ⁴C₁×⁴⁸C₄

= 4 × 1,94,580

= 7,78,320

 

Q7. In how many ways can one select a cricket team of eleven from 17 players inwhich only 5 players can bowl if each cricket team of 11 must include exactly 4bowlers?

A.7. We are given 17 players of which 5 players can bowl and 17 – 5 = 12 can bat. But we need to select a team of 11 in which there are exactly 4 bowlers.

Hence, the required number of ways

=⁵C₄ (bowl) x ¹²C_(11-4) (bat)

=  $\frac{5!}{4!\left(5-4\right)!}$ ×  $\frac{12!}{7!\left(12-7\right)!}$

=  $\frac{\mathrm{5\prime }<br>4!}{4!\prime <br>1!}$ ×  $\frac{1\mathrm{2\prime }<br>1\mathrm{1\prime }<br>1\mathrm{0\prime }<br>\mathrm{9\prime }<br>\mathrm{8\prime }<br>7!}{7!\prime <br>5!}$

= 5 ×  $\frac{1\mathrm{2\prime }<br>1\mathrm{1\prime }<br>1\mathrm{0\prime }<br>\mathrm{9\prime }<br>8}{\mathrm{5\prime }<br>\mathrm{4\prime }<br>\mathrm{3\prime }<br>\mathrm{2\prime }<br>1}$

= 5 × 11 × 9 × 8

= 3960

 

Q8. A bag contains 5 black and 6 red balls. Determine the number of ways in which2 black and 3 red balls can be selected.

A.8. We are given 5 black and 6 red balls of which 2 black and 3 red balls can be selected.

Thus the required number of ways

= ^5C₂×⁶C₃

=  $\frac{5!}{2!\left(5-2\right)!}$ ×  $\frac{6!}{3!\left(6-3\right)!}$

=  $\frac{\mathrm{5\prime }<br>\mathrm{4\prime }<br>3!}{\mathrm{2\prime }<br>3!}$ ×  $\frac{\mathrm{6\prime }<br>\mathrm{5\prime }<br>\mathrm{4\prime }<br>3!}{3!\prime <br>\mathrm{3\prime }<br>2}$

= 10 × 20

= 200

 

Q9. In how many ways can a student choose a programme of 5 courses if 9 coursesare available and 2 specific courses are compulsory for every student?

A.9. A student can choose 5 out of 9 available courses. However 2 specific courses are made compulsory.

So, now a student has 3 choices out of the remaining 7 courses.

Therefore, the required number of ways

=⁷C₃

=  $\frac{7!}{3!\left(7-3\right)!}$

=  $\frac{\mathrm{7\prime }<br>\mathrm{6\prime }<br>\mathrm{5\prime }<br>4!}{\mathrm{3\prime }<br>\mathrm{2\prime }<br>4!}$

= 35

 

Miscellaneous

Q1. How many words, with or without meaning, each of 2 vowels and 3 consonantscan be formed from the letters of the word DAUGHTER ?

A.1. In the word DAUGHTER, there are 3 vowels, namely A, U, E and 5 consonants, namely D, G, H, T, R.

The number of ways of selecting 2 vowels out of 3

= ³C₂

=  $\frac{3!}{2!\left(3-2\right)!}$

=  $\frac{\mathrm{3\prime }<br>2!}{2!\prime <br>1!}$

= 3

The number of ways of selecting 3 consonants out of 5

= ⁵C₃

=  $\frac{5!}{3!\left(5-3\right)!}$

=  $\frac{\mathrm{5\prime }<br>\mathrm{4\prime }<br>3!}{3!\prime <br>2!}$

= 5 × 2

= 10

Therefore, the number of combination of 3 consonants and 2 vowels is 3 × 10 = 30.

Each of these 30 combinations has 5 letters which can be arranged among themselves in 5! Ways.

Therefore, the required numbers of different words is

= 30 × 5! = 30 × 5 × 4 × 3 × 2 × 1 = 3600

 

Q2. How many words, with or without meaning, can be formed using all the letters ofthe word EQUATION at a time so that the vowels and consonants occur together?

A.2. In the word EQUATION, there are vowels (E, U, A, I, O) and 3 consonants (Q, T, N).

Treating vowels as a whole as 1st object and consonants as a whole as 2nd object, we can have an arrangement of 2! = 2.

Similarly, arrangement within the vowels = 5! = 5 × 4 × 3 × 2 × 1 = 120

And arrangement within the consonants = 3! = 3 × 2 × 1 = 6

Therefore, total number of possible arrangement = 2 × 120 × 6 = 1440

 

Q3. A committee of 7 has to be formed from 9 boys and 4 girls. In how many wayscan this be done when the committee consists of:

(i) exactly 3 girls ? (ii) atleast 3 girls ? (iii) atmost 3 girls ?

A.3. Given, a committee of 7 to be formed from 9 boys and 4 girls.

i.Since the committee will have exactly 3 girls therefore we can select 3 member from the given 4 girls and the remaining 4 member from the given 9 boys. Therefore, the required number of ways = ⁴C₃×⁹C₄

=  $\frac{4!}{3!\left(4-3\right)!}$ ×  $\frac{9!}{4!\left(9-4\right)!}$

=  $\frac{\mathrm{4\prime } <br>3!}{3!\prime <br><br>1!}$ ×  $\frac{\mathrm{9\prime }<br>\mathrm{8\prime }<br>\mathrm{7\prime }<br>\mathrm{6\prime }<br>5!}{\mathrm{4\prime }<br>\mathrm{3\prime }<br>\mathrm{2\prime }<br>\mathrm{1\prime }<br>5!}$

= 4 × 9 × 2 × 7

= 504

ii.Since, at least 3 girls are to be in the committee we can have the options.

(a) 3 girls and 4 boys selected in ⁴C₃×⁹C₄ ways

(b) 4 girls and 3 boys selected in ⁴C₄×⁹C₃ ways

Therefore the required number of ways.

= ⁴C₃×⁹C₄ + ⁴C₄×⁹C₃

=  $\frac{4!}{3!\left(4-3\right)!}$ ×  $\frac{9!}{4!\left(9-4\right)!}$ +  $\frac{4!}{4!\left(4-4\right)!}$ ×  $\frac{9!}{3!\left(9-3\right)!}$

=  $\frac{\mathrm{4\prime } <br>3!}{3!\prime <br>1!}$ ×  $\frac{\mathrm{9\prime }<br>\mathrm{8\prime }<br>\mathrm{7\prime }<br>\mathrm{6\prime }<br>5!}{\mathrm{4\prime } <br>\mathrm{3\prime }<br>\mathrm{2\prime }<br>\mathrm{1\prime }<br>5!}$ +  $\frac{4!}{4!}$ ×  $\frac{\mathrm{9\prime }<br>\mathrm{8\prime }<br>\mathrm{7\prime }<br>6!}{\mathrm{3\prime } <br>\mathrm{2\prime }<br>\mathrm{1\prime }<br>6!}$

= (4 × 9 × 2 × 7) + (3 × 4 × 7)

= 504 + 84

= 588

iii.Since, at the committee has to have at most 3 girls we can have the options.

(a) 0 girls and 7 boy selected in⁴C₀×⁹C₇ ways.

(b) 1 girl and 6 boys selected in ⁴C₁×⁹C₆ ways.

(c) 2 girls and 5 boys selected in ⁴C₂×⁹C₅ ways.

(d) 3 girls and 4 boys selected in ⁴C₃×⁹C₄ ways.

Therefore, the required number of ways.

= ⁴C₀×⁹C₇ + ⁴C₁×⁹C₆ + ⁴C₂×⁹C₅ + ⁴C₃ + ⁹C₄

=  $\frac{4!}{0!\left(4-0\right)!}$ ×  $\frac{9!}{7!\left(9-7\right)!}$ +  $\frac{4!}{1!\left(4-1\right)!}$ ×  $\frac{9!}{6!\left(9-6\right)!}$ +  $\frac{4!}{2!\left(4-2\right)!}$ ×  $\frac{9!}{5!\left(9-5\right)!}$ +  $\frac{4!}{3!\left(4-3\right)!}$ ×  $\frac{9!}{4!\left(9-4\right)!}$

=  $\frac{4!}{0}$ ×  $\frac{\mathrm{9\prime }<br>\mathrm{8\prime }<br>7!}{7!\prime <br>2!}$ +  $\frac{\mathrm{4\prime }<br>3!}{3!}$ ×  $\frac{\mathrm{9\prime } <br>\mathrm{8\prime }<br>\mathrm{7\prime }<br>6!}{6!\prime <br>3!}$ +  $\frac{\mathrm{4\prime } <br>\mathrm{3\prime }<br>2!}{\mathrm{2\prime } <br>2!}$ ×  $\frac{\mathrm{9\prime }<br>\mathrm{8\prime }<br>\mathrm{7\prime }<br>\mathrm{6\prime }<br>5!}{5!\prime <br>4!}$ +  $\frac{\mathrm{4\prime } <br>3!}{3!\prime <br>1!}$ ×  $\frac{\mathrm{9\prime }<br>\mathrm{8\prime }<br>\mathrm{7\prime }<br>\mathrm{6\prime }<br>5!}{\mathrm{4\prime }<br>\mathrm{3\prime }<br>\mathrm{2\prime }<br>\mathrm{1\prime }<br>5!}$

= (1 ×9 × 4) +  $\left(\mathrm{4\prime }\times \frac{\mathrm{9\prime }<br>\mathrm{8\prime }<br>\mathrm{7Ö}}{\mathrm{3\prime }<br>\mathrm{2\prime }<br>\mathrm{1Ö}}\right)$ +  $\left(\mathrm{2\prime }<br>\mathrm{3\prime }<br>\frac{\mathrm{9\prime }<br>\mathrm{8\prime }<br>\mathrm{7\prime }<br>\mathrm{6Ö}}{\mathrm{4\prime }3\prime <br>\mathrm{2\prime }<br>\mathrm{1Ö}}\right)$ + (4 × 9 × 2 × 7)

= 36 + 336 + 756 + 504

= 1632

 

Q4. If the different permutations of all the letter of the word EXAMINATION arelisted as in a dictionary, how many words are there in this list before the firstword starting with E ?

A.4. In the eleven-letter word EXAMINATION there are 2A’s, 2I’s and 2N’s and the rest are all different.

Since in a dictionary the words are listed according to the English alphabet we can only find words starting with A(as B, C, D are not a part of the letters forming the word EXAMINATION) listed before E.

Hence after fixing one A as first word we can rearrange the remaining 10 letters of which 2 are I, 2 are N and rest are all different.

Therefore, the required number of ways =  $\frac{10!}{2! 2!}$

=  $\frac{1\mathrm{0\prime }<br>\mathrm{9\prime }<br>\mathrm{8\prime }<br>\mathrm{7\prime }<br>\mathrm{6\prime }<br>\mathrm{5\prime }<br>\mathrm{4\prime }<br>\mathrm{3\prime }<br>2!}{\mathrm{2\prime } <br>2!}$

= 10 × 9 × 8 × 7 × 6 × 5 × 2 × 3

= 9,07,200

 

Q5. How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9which are divisible by 10 and no digit is repeated ?

A.5. Since the 6-digit numbers to be formed from the digits 0, 1, 3, 5, 7 and 9 has to be divisible by 10 we have to fix the unit place as 0. Now, the remaining 5 places can be filled only by the digits 1, 3, 5, 7 and 9.

Therefore, the required number of ways

= 5!

= 5 × 4 × 3 × 2 × 1

= 120

 

Q6. The English alphabet has 5 vowels and 21 consonants. How many words withtwo different vowels and 2 different consonants can be formed from thealphabet?

A.6. In an English word there are 5 vowels and 21 consonants.

The number of ways of selecting 2 vowel out of 5 = ⁵C₂

=  $\frac{5!}{2!\left(5-2\right)!}$

=  $\frac{\mathrm{5\prime }<br>\mathrm{4\prime }<br>3!}{\mathrm{2\prime } <br>\mathrm{1\prime }<br>3!}$

= 5 × 2 = 10

The number of ways of selecting 2 consonants out of 21 = ²¹C₂

=  $\frac{21!}{2!\left(21-2\right)!}$

=  $\frac{2\mathrm{1\prime } <br>2\mathrm{0\prime }<br>19!}{\mathrm{2\prime } <br>\mathrm{1\prime }<br>19!}$

= 21 × 10

= 210

Therefore, the number of combinations of 2 vowels and 2 consonants is 10 × 210 = 2100

Each of these 2100 combinations has 4 letters which can be rearranged among themselves in 4! Ways.

Therefore, the required number of ways

= 4! × 2100

= 4 × 3 × 2 × 1 × 2100

= 50400

 

Q7. In an examination, a question paper consists of 12 questions divided into two parts i.e., Part I and Part II, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions?

A.7. Since out of 8 total questions at least 3 questions has to be attempted from each of part I and II containing 5 and 7 questions respectively we can have the choices.

(a) 3 questions from I and 5 questions from II selected in ⁵C₃×⁷C₅ ways.

(b) 4 questions from I and 4 questions from II selected in ⁵C₄×⁷C₄ ways.

(c) 5 questions from I and 3 questions from II selected in ⁵C₅×⁷C₃ ways.

Therefore, the required number of ways.

= (⁵C₃×⁷C₅) + (⁵C₄×⁷C₄) + (⁵C₅×⁷C₃)

=  $\frac{5!}{3!\left(5-3\right)!}$ ×  $\frac{7!}{5!\left(7-5\right)!}$ +  $\frac{5!}{4!\left(5-4\right)!}$ ×  $\frac{7!}{4!\left(7-4\right)!}$ +  $\frac{5!}{5!\left(5-5\right)!}$ ×  $\frac{7!}{3!\left(7-3\right)!}$

=  $\frac{\mathrm{5\prime }<br>\mathrm{4\prime }<br>3!}{3!\prime <br>2!}$ ×  $\frac{\mathrm{7\prime } <br>\mathrm{6\prime }<br>5!}{5!\prime <br>2!}$ +  $\frac{\mathrm{5\prime }<br>4!}{4!\prime <br>1!}$ ×  $\frac{\mathrm{7\prime }<br>\mathrm{6\prime }<br>\mathrm{5\prime }<br>4!}{4!\prime <br>3!}$ +  $\frac{5!}{5!\prime <br>0!}$ ×  $\frac{\mathrm{7\prime }<br>\mathrm{6\prime }<br>\mathrm{5\prime }<br>4!}{\mathrm{3\prime }<br>\mathrm{2\prime }<br>\mathrm{1\prime }<br>4!}$

=  $\frac{\mathrm{5\prime }<br>4}{\mathrm{2\prime } <br>1}$ ×  $\frac{\mathrm{7\prime } <br>6}{\mathrm{2\prime }<br>1}$ +  $\frac{5}{1}$ ×  $\frac{\mathrm{7\prime }<br>\mathrm{6\prime }<br>5}{\mathrm{3\prime }<br>\mathrm{2\prime }<br>1}$ + 1 × 35

= (10 × 21) + (5 × 35) + 35

= 210 + 175 + 35

= 420

 

Q8. Determine the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.

A.8. In a deck of 52 card there are four kings.

So, number of ways of selecting exactly one king is ⁴C_1.

Now, after fixing one king card, we need to have the remaining 4 out of 5 cards to be a non-king i.e., only from the other 48 cards. So, number of ways of selecting is ⁴⁸C₄

Therefore, the required number of ways

= ⁴C₁×⁴⁸C₄

 

Q.9. It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?

A.9. As out of the total 9 seats 4 women are to be at even places we can have the following arrangement.

           

Seat places

	M	W	M	W	M	W	M	W	M
Seat places	1st	2nd	3rd	4th	5th	6th	7th	8th	9th

Also from this arrangement the women and men can rearrange among themselves.

Therefore, the required number of ways = 4! × 5!

= (4 × 3 × 2 × 1) × (5 × 4 × 3 × 2 × 1)

= 24 × 120

= 2880

 

Q10. From a class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can the excursion party be chosen?

A.10. In a class of 25 students, 10 students are to be selected for excursion. As 3 students decided that either all of them will join or none of them will join we have the options:

For the 3 students to be selected along with 7 other students from the remaining 25 – 3 = 22 students. This can be done in ³C₃ײ²C₇ ways.

For the 3 students to not be selected so that all 10 students will be from the remaining 25 – 3 = 22 students. This can be done in ³C₀ײ²C₁₀ ways.

Therefore, the required number of ways

= ³C₃× ²²C₇ + ³C₀ײ²C₁₀

= ²²C₇ + ²²C₁₀

 

Q11. In how many ways can the letters of the word ASSASSINATION be arranged so that all the S’s are together?

A.11. In the 13 letter word ASSASSINATION there are 3-A, 4-S, 2-I, 2-N, 1-T and 1-O.

Since all the S are to be occurred together we treat them i.e. (SSSS) as single object. This single object together with 13 – 4 = 9 remaining object will account for 10 objects having 3-A, 2-I, 2-N, 1-T and 1-O and can be rearranged in  $\frac{10!}{3!2!2!}$

=  $\frac{1\mathrm{0\prime }<br>\mathrm{9\prime }<br>\mathrm{8\prime }<br>\mathrm{7\prime }<br>\mathrm{6\prime }<br>\mathrm{5\prime }<br>\mathrm{4\prime }<br>3!}{3!\prime <br>\mathrm{2\prime }<br>1\mathrm{2\prime }<br>1}$

= 10 × 9 × 8 × 7 × 6 × 5

= 151200