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Permutation & combination

Questions



Q1. 33! is divisible by _____________________ .

(a) 233 (b) 241 (c) 230 (d) 239

Solution.
(c) 230

E2(33!) = E2(1.2.3.4------------.32.33)

= 16 + E2 (1.2.3---------.15.16)

= 16 + 8 + E2 (1.2.3.4------.8)

= 16 + 8 + 4 + E2(1.2.3.4) = 16 + 8 + 4 + 3 = 31

Note : 33! is divisible by 2n , n is integer having largest value 31 .
Solution

Q2. 8cr = 56

8pr = 336

find r .

Solution.
As we know that ncr . r! = npr
r! =
npr /ncr
=
336/56
= 6

r! = 6

r = 3
Solution

Q3. In how many way a group of 5 memebers can be developed from 6 males and 5 females consisting of 3 males and 2 females .

Solution.
3 males out of 6 can be selected 6c3 and 2 females can be selected 5c2

The required solution is 6c3 . 5c2 = 200 ways .
Solution

Q4. If 15c8 + 2 15c7 + 15c4 = 17cx find x.

Solution.
= ( 15c8 + 15c7) + ( 15c7 + 15c6)
= 16c8 + 16c7
= 17c8
x = 8
Solution

Q5. 8 women and 4 men are going to sit in a row so that men will occupy the even places . How many such arrangements are possible .

Solution.
In this case all 12 persons are to be seated in a row.

2nd , 4th , 6th , 8th , 10th and 12th are even places which is filled by 4 men in 6 p4 ways . 8 women will cover remaining 8 places 8p8 ways so , all possible arrangement are

6 p4 . 8p8
Solution

Q6. Find number of ways in which 7 men can be arranged in a row so that three particular men are seating together .

Solution.
Three persons are seating together so these three persons will be consider as a single person = 5! . 3!

Note : All three persons can interchange their position .
Solution

Q7. In how many ways can a party of 5 men and 5 women be seated at a circular table, so that no two women are adjacent ?

Solution.
First men will sit in 4! way , and five women can sit in 5! ways .
So total arrangements are = 4! . 5!
Note : In circular n person can sit in (n - 1)!
Solution

Q8. There are 12 points in a plane , in which 4 points are collinear, find how many straight lines can be drawn .

Solution.
Number of line formed by 12 points 12c2 number of straight lines from 4 collinear points 4c2 . All four collinear points develop a straight line .

The required answer is 12c2 - 4c2 + 1
Solution

Q9. How many 3 digits even numbers can be formed using the digits 0,1,2,3,4,5 (when repetition is not allowed)

Solution.
As we know, in even number unit place should be even .

(a) number with 4 at unit place 4 x 4 x 1 = 16

(b) number with 2 at unit place 4 x 4 x 1 = 16

(c) number with 0 at unit place 4 x 5 x 1 = 20

Total even numbers of 3 digits are 52 .
Solution

Q10. In how many ways can a examination papers be permutated so that the least and the best papers are never together (marks secured).

Solution.
= Total arrangements - arrangements in whih both papers are simultaneous

= 9p9 - 8p8 . 2! = 9! - 2.8! = 9.8! - 2.8!

= 8!(9 - 2) = 7.8 !

Note : The papers which are simultaneous consider as single paper .
Solution