Question

Find the value of
i) $\rm ^{20}C_{4} +\rm ^{20}C_{5}$
ii) $\rm ^{31}C_{26} - \rm ^{30}C_{26}$

Answer 1

Answer:

i) 20349

ii) 142506

Step-by-step explanation:

We know the formula of combination as shown below

nCr =  n! / r! (n - r)!

Therefore,

i)  20C4 + 20C5.

so, 20!/(4! * 16!) + 20!/(5! * 15!)

or, (20*19*18*17)/4! + (20*19*18*17*16)/5!

or, 4845 + 15504

or, 20349 [Ans]

ii)31C26 - 30C26

or, 31!/(26! * 5!) - 30!/(26! * 4!)

or, (31*30*29*28*27)/5! - (30*29*28*27)/4!

or, 169911 - 27405

or, 142506. [Ans]

Thus by applying a small formula you can easily solve such sums of permutation and combination.

Answer 2

Answer:

i) ²¹C₅

ii) ²⁰C₂₅

Step-by-step explanation:

Hi,

Consider ⁿCₓ₋₁ + ⁿCx

= n!/(x - 1)!(n-x+1)! + n!/(x)!(n-x)!

= n!/(x)!(n-x+1)![x + n - x + 1]

= n!/(x)!(n-x+1)![n + 1]

= (n + 1)!/x!(n+1-x)!

= ⁿ⁺¹Cₓ

Thus, ⁿCₓ₋₁ + ⁿCx = ⁿ⁺¹Cₓ

i) ²⁰C₄ + ²⁰C₅

Put n = 20 and x = 5 in ⁿCₓ₋₁ + ⁿCx = ⁿ⁺¹Cₓ , we get

= ²¹C₅

ii) ³¹C₂₆ - ³⁰C₂₆

Using ⁿCₓ₋₁ + ⁿCx = ⁿ⁺¹Cₓ

ⁿ⁺¹Cₓ - ⁿCx = ⁿCₓ₋₁ ,

Put n = 31, x = 26 we get

³⁰C₂₅

Hope, it helps !