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Chapter 7 Permutations and Combinations Ex 7.2 Solutions

Question - 1 : -
Evaluate
(i) 8!
(ii) 4! – 3!

Answer - 1 : -

(i) Consider 8!
We know that 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
= 40320
(ii) Consider 4!-3!
4!-3! = (4 × 3!) – 3!
Above equation can be written as
= 3! (4-1)
= 3 × 2 × 1 × 3
= 18

Question - 2 : - Is 3! + 4! = 7!?

Answer - 2 : -

Consider LHS 3! + 4!
Computing left hand side, we get
3! + 4! = (3 × 2 × 1) + (4 × 3 × 2 × 1)
= 6 + 24
= 30
Again consider RHS and computing we get
7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040
Therefore LHS ≠ RHS
Therefore 3! + 4! ≠ 7!

Question - 3 : - Compute 

Answer - 3 : -


Question - 4 : - If  find x.

Answer - 4 : -


Question - 5 : -
Evaluate
  ,
When
(i) n = 6, r = 2
(ii) n = 9, r = 5

Answer - 5 : -


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