Q19 of 118 Page 18

Evaluate

Given that we need to find the value of .


⇒


We know that nCr + nCr + 1 = n + 1Cr + 1


⇒


⇒


⇒


⇒


⇒


We know that


And also n! = n(n – 1)(n – 2)…………2.1


⇒


⇒


⇒


⇒


∴ The value of is 42504.


More from this chapter

All 118 →
17

For all positive integers n, show that

18

Prove that : 4nC2n : 2nCn = [1 . 3 . 5 …. (4n – 1)] : [1.3.5….. (2n – 1)]2.

 20

Let r and n be positive integers such that 1 ≤ r ≤ n. Then prove the following:

 20

Let r and n be positive integers such that 1 ≤ r ≤ n. Then prove the following:

n.n – 1Cr – 1 = (n – r + 1)(nCr – 1)