Question

8 different toys are to be placed in the boxes A and B.find the no of ways in which 8 toys can be placed in the boxes so that 5 toys are in  box A and 3 in box B.

Answer 1

First we have to select 5 toys from 8 toys in ⁸c₅ ways and then placed them in box A which can be done in 5! ways.

So, total no of ways of placing 5 toys out of 8 toys in box A is ⁸c₅ x 5! ways.

Now, we have left 3 toys which have to be placed in box B which can be done in 3! ways.

So, required number of ways 

= ⁸c₅ x 5! + 3!

= 6720 + 6

= 6726