Question

Combinatorics Proofs help
@ganeshie8

Answer 1

@ganeshie8

Answer 2

@ikram002p @perl @ParthKohli

Answer 3

@TheSmartOne

Answer 4

Sorry, I don't know :/

Answer 5

irs ok

Answer 6

@robtobey @Compassionate

Answer 7

@robtobey @Compassionate

Answer 8

@Abhisar

Answer 9

choose one

Answer 10

first one please

Answer 11

can u do the third one

Answer 12

@perl i completely dont understand how to do this type of question can u help

Answer 13

you want to show that \[\left {1}{b} \right _3^4 \]

Answer 14

\[\large ^{n+2}C_r =^{n}C_{r}+2*^{n}C_{r-1} +^nC_{r-2}\]

Answer 15

Part (a) asks you to prove (n choose r) + (n choose r + 1) = (n + 1 choose r + 1).

One strategy is to rewrite the function (n choose k) as n!/(k!*(n-k)!), and show that both sides of the equality match.

So, rewriting the LHS first:

(n choose r) + (n choose r + 1) = n!/r!(n-r)! + n!/(r+1)!/(n-r-1)! =
n!(r+1)+n!(n-r)! / (r+1)!(n-r)! = n!(n+1)/(r+1)!(n-r)! = (n+1)!/(r+1)!(n-r!)

This is just (n + 1 choose r + 1), thus we have shown that (n choose r) + (n choose r + 1) = (n + 1 choose r + 1).

Answer 16

I made a typo.

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

Answer 17

So, it amounts to substituting the (n choose k)'s and then using algebraic manipulations to demonstrate equality.

Answer 18

look at step

Answer 19

that would be a feat to write that in Latex