 # Important Questions For Class 11 Maths- Chapter 7- Permutation and Combination

Permutation and Combination is one of the crucial topics of Maths. It is an important topic for entrance exams like JEE and also for class 11(CBSE). Apart from studying and practicing problems on permutation and combination from NCERT, students shall also practice these important questions.

Solving these important questions of class 11 maths chapter 7 will help you prepare for joint entrance exams too.

Q.1: Find the 2-digit numbers that can be formed from the given digits:

1, 2, 3, 4 and 5

assuming that

a) digits can be repeated

b) digits are not allowed to be repeated.

Soln:

a) By the multiplication principle, the number of ways in which three-digit numbers can be formed from the given digits is 5 × 5 × 5 = 125

b) By the multiplication principle, the number of ways in which three-digit numbers can be formed without repeating the given digits is 5 × 4 × 3 = 60

Q.2: A coin is tossed 6 times and the outcomes are noted. How many possible outcomes can be there?

Soln : When we toss a coin once, the number of outcomes we get is 2 (Either Head or tail)

So, in each throw, the no. of ways to get a different face will be 2.

Therefore,, by multiplication principle, the required no. of possible outcomes is

2 x 2 x 2 x 2 × 2 × 2 = 64

Q.3: Evaluate the following

(i) 6 ! (ii) 5 ! – 2 !

Soln :

(i) 6! = 1 × 2 × 3 × 4 × 5 × 6 = 720

(ii) 5! = 1 × 2 × 3 × 4 x 5 = 120

As 2! = 1 × 2 = 2

Therefore, 5! – 2! = 120 – 2 = 118

Q.4: From a team of 6 students, in how many ways can we choose a captain and vice captain assuming one person can not hold more than one position?

Soln : From a team of 6 students, two students are to be chosen in such a way that one student will hold only one position.

Here, the no. of ways of choosing a captain and vice captain is the permutation of 6 different things taken 2 at a time.

So, 6P2 = 6! / ( 6 -2 )! = 6! / 4! = 2! = 2

Q5. How many words can be formed using all the letters of the word Dinomite, using each letter exactly one time?

Soln: As there are 7 different letters in the word – Dinomite. So the number of different words formed using these 7 letters will be 8p7 = 8! / (8 -7)! = 8

Q.6: How many words can be formed each of 2 vowels and 3 consonants from the letters of the given word – DAUGHTER ?

Soln : No. of Vowels in the word – DAUGHTER are 3.

No. of Consonants in the word Daughter are 5.

No of ways to select a vowel = 3c2 = 3!/(3 – 2)! = 3

No. of ways to select a consonant = 5c3 = 5!/(5 – 3)! = 10

Now you know that the number of combinations of 3 consonants and 2 vowels = 10 x 3 = 30

And these can be arranged in 30 x 5! = 3600 ways

Q.7: It is needed to seat 5 boys and 4 girls in a row so that the girl get the even places. How many such arrangements are possible ?

Soln : 5 boys and 4 girls are to be seated in a row so that the girl gets the even places.

The 5 boys can be seated in 5! ways.

For each of the arrangement, the 4 girls can be seated only at the places which are cross marked to make girls occupy the even places).

B x B x B x B x B

So, the girls can be seated in 4! ways.

Hence, the possible number of arrangements = 4! × 5! = 24 × 120 = 2880

Q.8: Find the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.

Soln : Take a deck of 52 cards,

To get exactly one king, a 5-card combinations have to be made. It should be made in such a way that in each selection of 5 cards,or in a deck of 52 cards, there will be 4 kings.

To select 1 king out of 4 kings = 4c1

To select 4 cards out of remaining 48 cards = 48c4

To get the needed number of 5 card combination = 4c1 x 48c4 =

Q9. Find the number of 6 digit numbers that can be formed by using the digits 0, 1, 3, 5, 7, and 9. These digits shall be divisible by 10 and no digit shall be repeated?

Soln: The number which has a 0 in its units place is divisible by 10.

If we put 0 in the unit place, _ _ _ _ 0, there will be as many ways to fill 5 vacant places. (1, 3, 5, 7, 9)

The five vacant places can be filled in 5! ways = 120.

Q10. Evaluate: 10! – 6!

Soln: 10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x1

6! = 6 X 5 x 4 x 3 x 2 x 1

10! – 6! = 10 x 9 x 8 x 7 = 5040