Test: Permutation of Distinct Objects - JEE MCQ

3 executives in 4 chairs = 4P3 ways
Total no. of ways = 2! * 4P3
=> 2*1*(4!/1!)
= 2*24
= 48

Let the friends be A,B,C,D,E,F,G,H,I,J and assume A and B will not battend together
Case 1 : Both of them will not attend the party : Now we have to select the 6 guests from the remaing 8 members 
Then no of ways is ⁸C₆ = 28 ways.
Case 2 : Either of them are selected for the party :Now we have to select the 5 guests from the remaing 8 members and  one from A and B
Then the no of ways = ²C₁ x ⁸C₅ =112 ways.
Therefore total number of ways is 28+ 112 = 140 ways,

Excluding the 2 tallest boys, we can divide the 8 boys into 2 groups of 4 each by
8!/4!4!2! ways.
The tallest boys can be assigned to the groups in 2 ways.
The desired number of ways are

The total no. of ways of dividing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is:

The number of ways =21.