How to prove: $^nC_r + ^nC_{r+1} = ^{n+1}C_{r+1}$

Question

Prove: ${^n}C_r + ^nC_{r+1} = ^{n+1}C_{r+1}$

I used $^nC_r = \frac{(n!)}{(n-r)!r!}$ and then did my substitutions:

$ = \frac{(n!)}{(n-r)!r!} + \frac{(n!)}{(n-(r+1))!(r+1)!}$

$ = \frac{(n!)}{(n-r)!r!} + \frac{(n!)}{(n-r-1)!(r+1)!}$

$=$

I am not sure how to do rest of the working out

Well, you have two fractions that you want to add together to a single fraction (since you want to compare it to $^{n+1}C_{r + 1}$). How would you usually do that? — Arthur, Feb 28 '19 at 14:44
Does this answer your question? Proving Pascal's Rule : ${{n} \choose {r}}={{n-1} \choose {r-1}}+{{n-1} \choose r}$ when $1\leq r\leq n$. Your $r$ and $n$ values are $1$ less than in the other question, and different notation is used, but otherwise they're the same, with this being called as commented in the other question, Pascal's rule. — John Omielan, Aug 17 '21 at 05:36

Abhinav · Answer 1 · 2019-02-28T14:53:21.767

0

$$\frac{n!\cdot{(r+1)} +n!\cdot{(n-r)}}{(n-r)!\cdot{(r+1)!}}$$ $$=\frac{n!\cdot{(n+1)}}{((n+1)-(r+1))!\cdot{(r+1)!}} $$ $$=^{n+1}C_{r+1}$$

edited Feb 28 '19 at 14:53

answered Feb 28 '19 at 14:47

Abhinav

450

how did you get $${(n-r)!\cdot{(r+1)!}}$$ – Dev Patel Feb 28 '19 at 15:05
See the denominator – Abhinav Feb 28 '19 at 15:07
In the denominator we take the LCM – Abhinav Mar 01 '19 at 03:21

score 0 · Answer 2 · answered Mar 01 '19 at 15:08

A combinatorical way: Choose $r+1$ elements from a set of $n+1$ different element. This can be done in $\dbinom{n+1}{r+1}$ or we can count the ways to choose $r+1$ elements from the same set without the element $X$ and the ways to choose $r+1$ elements from the set where $X$ is part of these $r+1$ numbers, this can be done in $\dbinom{n}{r+1}+\dbinom{n}{r}$ . And since we are counting the same thing we are done.

user · Answer 3 · 2019-02-28T17:26:43.533

-1

$$\begin{align} \binom n{k+1}+\binom n{k}&=\frac{n!}{(k+1)!(n-k-1)!}+\frac{n!}{k!(n-k)!}\\ &=\frac{n!}{(k+1)!(n-k-1)!}\left[1+\frac{k+1}{n-k}\right]\\ &=\frac{n!}{(k+1)!(n-k-1)!}\cdot\frac{n+1}{n-k}\\ &=\frac{(n+1)!}{(k+1)!(n-k)!}=\binom{n+1}{k+1}. \end{align} $$

edited Feb 28 '19 at 17:26

answered Feb 28 '19 at 14:46

user

26,272

how does that equal to (n+1 K+1) – Dev Patel Feb 28 '19 at 14:57
Just replace $k$ with $k+1$ in both sides of equality. – user Feb 28 '19 at 14:59
can you please show those changes on your solution – Dev Patel Feb 28 '19 at 15:01
@DevPatel Done. – user Feb 28 '19 at 15:10
what happens to the (n-k)! – Dev Patel Feb 28 '19 at 15:12
@DevPatel Which one? – user Feb 28 '19 at 15:13
the answer u have given is (n+1 K+1) but before that step what happened to (n-k)! in $\frac{(n+1)!}{(k+1)!(n-k)!}$ – Dev Patel Feb 28 '19 at 15:16
1

$n-k=(n+1)-(k+1).$ – user Feb 28 '19 at 15:18
Let us continue this discussion in chat. – Dev Patel Feb 28 '19 at 15:20
The downvoter is welcome to express his/her motivation. – user Oct 10 '21 at 10:48

How to prove: $^nC_r + ^nC_{r+1} = ^{n+1}C_{r+1}$

3 Answers3