Question

Evaluate 50c11+50c12+51c13-52c13

Answer 1

Answer:

The value of $50C_{11}+50C_{12}+51C_{13}-52C_{13}$

is 0

Step-by-step explanation:

Formula used:

$nC_r=$ Number of combinations of n things taken r at a time.

$nC_r+nC_{r-1}=(n+1)C_r.........(1)$

$Now,\\\\50C_{11}+50C_{12}+51C_{13}-52C_{13}\\\\=51C_{12}+51C_{13}-52C_{13}\\\\=52C_{13}-52C_{13}\\\\=0$

Answer 2

Answer:

0

Step-by-step explanation:

Evaluate 50c11+50c12+51c13-52c13

⁵⁰C₁₁ + ⁵⁰C₁₂ + ⁵¹C₁₃ - ⁵²C₁₃

ⁿCₓ =  n! / (x!) (n-x)!

= 50!/( 11! * 39! )  + 50!/(12! * 38!) + 51!/(13! * 38!)  - 52!/ (13! * 39!)

=  50!/(11! * 38 !) (  1/39 + 1/12  +  51/(13*12)  - 52*51/(13 * 12 * 39)

=  50!/(11! * 38 !) (1/(39 * 12 * 13) ( (12 * 13)  + (39 * 13)  + (51*39)   - 52*51))

= 50!/(11! * 38 !) (1/(39 * 12 * 13) (156 + 507 + 1989 - 2652)

= 50!/(11! * 38 !) (1/(39 * 12 * 13) (2652 - 2652)

= 50!/(11! * 38 !) (1/(39 * 12 * 13) (0)

= 0