Q1.
There are 10 students in a class including 4 girls . The number of ways to arrange them in a row when any two girls out of four never
comes together .
Solution.
Let first 6 boys will sit in 6! ways
Now there are 7 places to cover by 4 girls in 7p4 ways .
6! . 7p4
Solution
Q2.
Six points are on a circle . The number of quardrilaterals that can be developed .
Solution
A quardrilaterals can be developed joining by four points . So selecting four points out of 6 points 6c4 .
So number of quadrilateral are 6c4 = 15
Solution
Q3.
If npr = 720 and ncr = 120, find the value of r .
Solution.
nc
r . r! =
np
r r! =
npr/ncr
=
720/120
= 6
r! = 3!
r = 3
Solution
Q4.
If 15c8 + 2 15c7 + 15c4 = 17cx find x.
Solution.
= ( 15c8 + 15c7) + ( 15c7 + 15c6)
= 16c8 + 16c7
= 17c8
x = 8
Solution
Q5.
(n + 2)! = 72n! find n .
Solution.
(n + 2)(n + 1)n! = 72 n!
(n + 2)(n + 1) = 72
n = 7
Solution
Q6.
769c60 = 768cx + 768c60 , find x .
Solution.
x = 59
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.
In how many ways can a party of 4 men and 4 women be arranged at a circular table , so that no two men are adjacent .
Solution.
= 3! . 4! = 144
Solution
Q9.
In how many way can 14 billiard balls be arranged if 7 of them are red and 7 of them are green .
Solution
Q10.
20cr = 20cr + 4 , find 8cr .
Solution.
r + r + 4 = 20
2r = 16
r = 8
8cr = 8c8 = 1
Solution
