Difference between revisions of "A choose b"

Revision as of 19:26, 15 June 2019

Here is the formula for a choose b: $\binom{a}{b}=\frac{a!}{b!(a-b)!}$ . This is assuming that of course $a \ge b$ .

Why is it important?

a choose b counts the number of ways you can pick b things from a set of a things. For example $\binom{8}{2}=\frac{8!}{2!(8-2)!}=\frac{8(7)}{2}=\frac{42}{2}=21$ . More at https://artofproblemsolving.com/videos/counting/chapter4/64.

a choose 2

Here is a list of n choose 2's

$\binom{2}{2}=1$

$\binom{3}{2}=3$

$\binom{4}{2}=6$

$\binom{5}{2}=10$

These are triangle numbers! My proof uses induction (assuming something is true unless proofed true or not true). $\binom{n}{2}=1+2+3...+(n-1)$ Then Simplify:

$\frac{n!}{2!(n-2)!}=\frac{n(n+1)}{2}$ More Simplify:

$\frac{n(n+1)}{2}=\frac{n(n+1)}{2}$

So now we have proved it. If you don't get what I did on the second step go to Proof Without Words on this wiki.

Pascal's Identity

Pascal's Identity states that $\binom{n}{k}+\binom{n}{k+1}=\binom{n+1}{k+1}$ .

Why: $\binom{n}{k}+\binom{n}{k+1}=\binom{n+1}{k+1} \Rightarrow \frac{n!}{k!(n-k)!}+\frac{n!}{(k+1)!(n-(k+1)!}$

@@ Line 27: / Line 27: @@
 == Pascal's Identity ==
-Pascal's  Identity states that <math>\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}</math>
+Pascal's Identity states that <math>\binom{n}{k}+\binom{n}{k+1}=\binom{n+1}{k+1}</math>.
+Why: <math>\binom{n}{k}+\binom{n}{k+1}=\binom{n+1}{k+1} \Rightarrow \frac{n!}{k!(n-k)!}+\frac{n!}{(k+1)!(n-(k+1)!}</math>

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Difference between revisions of "A choose b"

Revision as of 19:26, 15 June 2019

Why is it important?

a choose 2

Pascal's Identity