In how many ways can 10 balls be divided between two boys, one receiving two and the other eight balls?

Open in App

Solution

When A receives 2 and B gets 8, then = $\frac{(10!)}{2! . 8!} = 45$
When A receives 8 and B gets 2, then also = $\frac{(10!)}{2! . 8!} = 45$
$∴$ Total is 45 + 45 = 90.

Suggest Corrections

Similar questions

Q. In how many ways can

10

different balls be divided between two boys, one receiving two and the other eight balls

Q. In how many ways can

10

balls be distributed to two boys, one receiving two and the other eight balls?

In how many ways can 12 balls be divided between 2 boys, one receiving 5 and the other 7 balls?

___

Q. The number of ways in which 12 balls can be divided between two friends, one receiving 8 and the other 4, is

There are 5 historical monuments, 6 gardens and 7 shopping malls in the city. In how many a tourist can visit the city, if he visits at least one shopping mall?

Join BYJU'S Learning Program

In how many ways can 10 balls be divided between two boys, one receiving two and the other eight balls?

When A receives 2 and B gets 8, then = (10!)2!.8!=45When A receives 8 and B gets 2, then also = (10!)2!.8!=45∴ Total is 45 + 45 = 90.

When A receives 2 and B gets 8, then = $\frac{(10!)}{2! . 8!} = 45$
When A receives 8 and B gets 2, then also = $\frac{(10!)}{2! . 8!} = 45$
$∴$ Total is 45 + 45 = 90.