Class 11 Maths Ncert Solutions Chapter 7 Ex 7.5 Permutations & Combinations PDF

# Class 11 Maths Ncert Solutions Ex 7.5

## Class 11 Maths Ncert Solutions Chapter 7 Ex 7.5

Q-1: Find the number of words, having meaning or meaningless, each of 2 vowels and 4 consonants will be formed by the letters of the word DAUGHTER?

Solution:

In DAUGHTER word, we have 3 vowels which is A, U and E, rest 5 letters are consonants which are D, G, H, T and R.

Now,

The number of way to select 2 vowels at a time from 3 vowels will have permutation say, 3C2 = 3

Also, the number of way to select 4 consonants at a time from 5 consonants will have permutation say, 5C4= 5

Hence, the total number of combinations of 2 vowels and 4 consonants = 3 × 5 = 15

These 15 combinations of 2 vowels and 4 consonants will be arranged in 6! number of ways.

Therefore, the number of different words required = 15 × 5! = 1800

Q-2: Find the number of words, which can be either meaningful or meaningless, which will be formed by using all of the letters of the word EQUATION at a time such that both the vowel and the consonant will occur together.

Solution:

In the given word EQUATION, we have 5 vowels, that are E, U, A, I and O, and there are 3 consonants, that are Q, T and N.

As all the vowels and consonants will occur together, means both (A, E, I, O, U) and (Q, T, N) will be considered as a single letter.

Thus, the permutation of vowels and consonants taken all at a single time is counted which is: 2P2 = 2!

Now, at each of the permutation for vowels, there is a permutation of 5! Taken all at a time and for consonants, the permutation is 3! taken all at a time.

Therefore, by the principle of multiplication, number of the words required = 2! × 5! × 3! = 1440

Q-3: There are 10 boys and 5 girls through which a committee with 7 members has to be formed. Now, find the number of ways in which this might be created when the committee formed consists of:

(i) Exactly 4 girls                          (ii) at least 4 girls                            (iii) almost 4 girls?

Solution:

(i) From 10 boys and 5 girls, a committee of 7 members is being formed.

As in this case, exactly 4 girls is needed to be there in the committee, so each committee should contain ( 7 – 4 ) = 3 boys in the committee.

Therefore, number of ways required in this case:

= 5C4 × 10C3

= 5!4!1!×10!3!7!$\frac{ 5! }{ 4!\; 1! } \times \frac{ 10! }{ 3!\; 7! }$ = 5×10×9×8×7!7!×3×2×1$5 \times \frac{ 10\; \times \;9\; \times \;8\; \times \;7! }{ 7!\; \times \;3\; \times \;2\; \times \;1 }$

= 600 number of ways

(ii) Here, in this case, at least 4 girls can be there in each committee.

So, in such cases, a committee can consist of

(a) Either 3 boys or 4 girls                                            (b) either 2 boys or 5 girls.

(a) 4 boys and 3 girls will be selected in 10C3 × 5C4 number of ways.

(b) 2 boys and 5 girls will be selected in 10C2 × 5C5 number of ways.

Hence, the number of ways required in this case:

= 10C3 × 5C4 + 10C2 × 5C5

=10!3!7!×5!4!1!+10!2!8!×5!5!0!$\frac{ 10! }{ 3!\; 7!} \times \frac{ 5! }{ 4! \;1! } + \frac{ 10! }{ 2! \;8! } \times \frac{ 5! }{ 5! \;0!}$

= [10×9×8×7!3×2×1×7!×5]+[10×9×8!2×1×8!×1]$\left [\frac{ 10\; \times \;9\; \times \;8\; \times\; 7! }{ 3\; \times \;2\; \times \;1\; \times \;7! } \times \;5\; \right ] + \left [ \frac{ 10\; \times \;9\; \times \;8! }{ 2\; \times \;1\; \times \;8! } \times 1 \right ]$= ( 10 × 3 × 4 × 5 ) + ( 5 × 9 )

= 645 number of ways.

(iii) Here, in this case, almost 4 girls can be there in each of the committee.

So, in this case, a committee can consist of:

(a) 4 girls and 3 boys                                                                (b) 3 girls and 4 boys

(c) 2 girls and 5 boys                                                                (d) 1 girl and 6 boys

(e) No girl and 7 boys

(a) 4 girls and 3 boys will be selected in 5C4 × 10C3 number of ways.

(b) 3 girls and 4 boys will be selected in 5C3 × 10C4 number of ways.

(c) 2 girls and 5 boys will be selected in 5C2 × 10C5 number of ways.

(d) 1 girl and 6 boys will be selected in 5C1 × 10C6 number of ways.

(e) No girl and 7 boys will be selected in 5C0 × 10C7 number of ways

Hence, the number of ways required in this case:

= 5C4 × 10C3 + 5C3 × 10C4 + 5C2 × 10C5 + 5C1 × 10C6 + 5C0 × 10C7

= 5!4!1!×10!3!7!+5!3!2!×10!4!6!+5!2!3!×10!5!5!+5!1!4!×10!6!4!+5!0!5!×10!7!3!$\frac{ 5! }{ 4! \;1!} \times \frac{ 10! }{ 3! \;7! } + \frac{ 5! }{ 3! \;2! } \times \frac{ 10! }{ 4!\; 6!} + \frac{ 5! }{ 2!\; 3! } \times \frac{ 10! }{ 5! \;5!} + \frac{ 5! }{ 1!\; 4! } \times \frac{ 10! }{ 6!\; 4!} + \frac{ 5! }{ 0! \;5! } \times \frac{ 10! }{ 7! \;3!}$

= 300 + 2100 + 2520 + 1050 + 120

= 6090 number of ways.

Q-4: In a dictionary, the permutation of the letters of the word EXAMINATION is listed, then find the total number of words in the list of dictionary before the first letter of the word start with E.

Solution:

There are 11 letters in the given word EXAMINATION where A, I and N appeared twice and all other letter appeared only for once.

Words which are listed before the word which is starting with E in a dictionary can be the word which starts with only A.

Hence, in order to get the numbers of words starting from A, means the letter A will be fixed at the extreme left position of the word, and there are 10 remaining letters taken at a time is rearranged.

In the given word which is EXAMINATION, there are 2 N’s and 2 I’s in the remaining letters and other letters occurred only for once.

Thus, the number of words starting with the letter A:

= 10!2!2!$\frac{ 10! }{ 2!\; 2! }$ = 907200

Hence, the number of words required is 907200.

Q-5: Find the total number of 6- digit numbers formed by using the digits 0, 2, 3, 4, 5 and 6 which should be divisible by 10. Note that, the repetition of digit is not allowed.

Solution:

For a number to be divisible by 10, then its unit place must be filled with 0 only.

Thus, units place is fixed which is to be filled only by 0.

Hence, there are multiple numbers of ways to fill up the other 5 vacant places in succession with the other digits that is, 2, 3, 4, 5 and 6.

Now,

Other 5 vacant places will be filled in by 5! number of ways.

Therefore, the number of 6-digit numbers required = 5! = 120

Q-6: It is known that the English alphabet has 21 consonants and 5 vowels. Find the number of words formed with 4 different consonants and 3 different vowels from the English alphabet.

Solution:

We need to select 4 different consonants from 21 consonants and 3 different vowels from 5 vowels of the English alphabet.

So, the permutation of the selection of vowels is given by:

5C3 = 5!3!2!$\frac{ 5! }{ 3!\; 2! }$ = 10

Also, the permutation of the selection of consonants is given by:

21C2= 21!4!17!$\frac{ 21! }{ 4! \;17! }$ = 3990

Hence, the total number of the combinations of 4 different consonants and 3 different consonants = 10 × 3990 = 39900.

Q-7: There are 12 questions in the question paper of an examination which is mainly divided into two parts, say, part-I and part-II, each having 4 and 8 questions, respectively. There is a condition given in the question paper that the student has to attempt at least 8 questions for sure, selection 3 from each section. Now, find the number of ways in which a student can select the questions in the question paper.

Solution:

From the data given in the question, know that the examination question paper is divided into two parts, namely; part-I and part-II, containing 4 and 8 questions each.

A student need to attempt at least 8 questions, selecting 3 questions at least from each sections, which can be done in following ways:

(a) 4 questions from part I and 4 questions from part II

(b) 3 questions from part I and 5 questions from part II

The selection of 4 questions from part I and 4 questions from part II have permutation as:  4C4 × 8C4 number of ways.

The selection of 3 questions from part I and 5 questions from part II have permutation as:  4C3 × 8C5 number of ways.

Now,

The total number of ways in which selection of questions are required:

= 4C4 × 8C4 + 4C3 × 8C5

=4!4!0!×8!4!4!$\frac{ 4! }{ 4!\; 0! } \times \frac{ 8! }{ 4!\; 4! }$ + 4!3!1!×8!5!3!$\frac{ 4! }{ 3!\; 1! } \times \frac{ 8! }{ 5! \;3! }$ = 1+8×7×6×5!5!×3×2×1$1 + \frac{ 8\; \times \;7\; \times \;6\; \times \;5! }{ 5!\; \times \;3\; \times \;2\; \times \;1 }$

= 57 number of ways

Q-8: Find the number of ways of 5- card combinations from a deck of 52 cards. Note that, among those 5 cards, there must be exactly one king in the combination.

Solution:

We need to make 5 – card combinations from a deck of 52 cards, in such a way that there must be a king in the combination.

We know that, there are 4 kings in a deck of 52 cards.

Then,

The permutation of the selection of a king from the deck is: 4C1 number of ways.

Now after selection of a king for the combination of 5 cards then, there are 4 cards left to be selected.

Thus, 4 cards can be selected from the deck with 48 cards left in 48C4 number of ways.

Therefore, the number of 5- card combination required is:

=4C1 × 48C4 = 4!1!3!×48!4!44!$\frac{ 4! }{ 1!\; 3! } \times \frac{ 48! }{ 4! \;44! }$ = 48×47×46×45×44!3×2×1×44!$\frac{ 48\; \times \;47\; \times \;46\; \times\;45\; \times \;44! }{ 3\; \times \;2\; \times \;1\; \times \;44! }$

= 7,78,320 number of combinations.

Q-9: Consider a situation in which 7 men and 6 women needs to be seated in a row in such a way that the women will occupy the even places.

Solution:

It is required to arrange the seating of 7 men and 6 women in a row in such a way that the women will occupy the even places.

7 men can be seated in 7! Ways. For every arrangement, 6 women will be seated only at the blank ( _ ) places so that the women can occupy the even places.

A _ A _ A _ A _ A _ A_ A

Thus, the women will be seated in 6! number of ways.

Therefore, Number of arrangements possible as per the given condition = 6! × 7! = 3628800 number of arrangements.

Q-10: There are 20 students in a class from which 8 students will be chosen for an exclusion party. 3 students from the class decided that either they will join combinely or none of them will go for the exclusion party. Find the number of ways by which the exclusion party will be chosen.

Solution:

8 students will be selected for the exclusion party from a class of 20 students.

Also, it is given in the question that, 3 students from that class decided that they will either go together or none of them will go for the exclusion party. Thus, there are mainly 2 cases:

(a) All 3 of them are going for the exclusion party.

Thus, the remaining 5 students will be selected from the remaining 17 students in the class in 17C5$^{ 17 }C_{ 5 }$ number of ways.

(b) None of them are going for the exclusion party.

Thus, 8 students will be selected from the class of 20 students in 20C8$^{ 20 }C_{ 8 }$ number of ways.

Therefore, the number of ways required for choosing students for the exclusion party is:

=17C5+20C8$^{ 17 }C_{ 5 } + \;^{ 20 }C_{ 8 }$ = 17!5!12!×20!8!12!$\frac{ 17! }{ 5! \;12! } \times \frac{ 20! }{ 8! \;12! }\\$

= 17×16×15×14×13×12!5×4×3×2×1×12!$\frac{ 17\; \times \;16\; \times \;15\; \times \;14\; \times \;13\; \times \;12! }{ 5\; \times \;4\; \times \;3\; \times \;2\; \times \;1\; \times \;12! }$ + 20×19×18×17×16×15×14×13×12!8×7×6×5×4×3×2×1×12!$\frac{ 20\; \times \;19\; \times \;18\; \times \;17\; \times \;16\; \times \;15\; \times \;14\; \times \;13\; \times \;12! }{ 8\; \times \;7\; \times \;6\; \times \;5\; \times \;4\; \times \;3\; \times \;2\; \times \;1\; \times 12! }$ = 6188 + 125970

= 132158 number of ways to choose the students for the exclusion party.

Q-11: Arrange ASSASSINATION in such a way that all the S’s are filled together. Find the number ways for the arrangement of S’s together.

Solution:

ASSASSINATION is the word given in the question where A appeared for 3 times, S appeared for 4 times, I appeared for 2 times, N appeared for 2 times and, T and O appeared just for once.

We need to arrange all the words in such a way that the all S’s are placed together.

Now,

SSSS is considered as a single object.

Since, including the single object we have remaining 9 letters, so we have the count of 10 letters more.

Thus, the 10 letters where they have 3 A’s, 2 I’s and 2 N’s will be arranged in 10!3!2!2!$\frac{ 10! }{ 3! \;2! \;2! }$ number of ways.

Therefore, the total number of ways required to arrange the letters of ASSASSINATION word = 10!3!2!2!$\frac{ 10! }{ 3! \;2! \;2! }$ = 151200