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,

nP0 
= n!/(n-0)! 
=> n!/n!  = 1

nPr = 10
nPr = n!/(n-r)!
10 = n!/(n-1)!
10 = [n!.(n-1)!]/(n-1)!
n! = 10

No. of ways of selecting two flags out of four = 4C2
So, total possible different signals generated =  4C2×2!
⟹ 6×2=12

nPn
= n!/(n-n)!
= n!

5P5  ⇒ 5!/(5-5)!
= 5!/0!
= 5!/1 ⇒ 5!
= 120 ways

The value of 0! = 1