Evaluate

- 8!
- 4! – 3!

Is 3! + 4! = 7!?

Compute

8!/(6!X 2!)

Calculate x if

Evaluate,

n!/(n-r)!

when

(i) n = 6, r = 2

(ii) n = 9, r = 5

We know that

n!=n(n-1)(n-2)……3*2*1

So solving the below questions based on that

- 8! = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 = 40320
- First calculate the factorials
4! = 1 × 2 × 3 × 4 = 24

3! = 1 × 2 × 3 = 6

Then 4! – 3! = 24 – 6 = 18

We know that factorials is given by

n!=n(n-1)(n-2)……3*2*1

So solving the below questions based on that

3! = 1 × 2 × 3 = 6

4! = 1 × 2 × 3 × 4 = 24

3! + 4! = 6 + 24 = 30

7! = 1 × 2 × 3 × 4 × 5 × 6 × 7 = 5040

So it is clear that

3! + 4! ≠ 7!

8!/(6!X 2!)

= 8*7 *6!/6! * 2

=8*7/2=28

Solving this

x=64

n!/(n-r)!

when

(i) n = 6, r = 2

(ii) n = 9, r = 5

n!/(n-r)!= 6!/4!=30

n!/(n-r)!= 9!/4!=9*8*7*6*5=15120

Download this assignment as pdf

**Notes**- What is factorial
- Fundamental Principle of counting
- Permutation
- Restricted Permutations
- Another Theorem of Permutation
- What is combination
- Combination Formula
- Properties of Combination Formula

**NCERT Solutions**