# A binomial identity

• chaotixmonjuish

#### chaotixmonjuish

$$\sumk=0n\binom{n}{k}2=\binom{2n}{n}$$

Could someone give me a hint as to how to start this. I'm not sure how to really interpret it.

$$(n-k)\binom{n}{k}=n\binom{n-1}{k}$$
Right Side: Suppose you create a committe from $$\binom{n}{k}$$, then to pick a leader who isn't in the committee but in the pool of people, we have n-k ways.

Left Side: Suppose you have n ways to pick a leader for a group. After selecting the leader, you have n-1 people left to pick a committee of size k.

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## Answers and Replies

Hi chaotixmonjuish !

(try using the X2 and X2 tags just above the Reply box )
k=0n nCk2 = 2nCn

Could someone give me a hint as to how to start this. I'm not sure how to really interpret it.

The RHS is the number of ways of choosing n people from 2n people.

Hint: Suppose the 2n people are n men and n women.

So would the right hand side be saying that suppose we had n men and n women, there are n ways to form a committee consisitng of both men and women.

So would the right hand side be saying that suppose we had n men and n women, there are n ways to form a committee consisitng of both men and women.

uhhh?

the RHS is the same number, no matter how many men (or women) there are.

Uh oh, ha ha, now I'm confused...I feel like this binomial identiy has some really easy RHS.

Does it just count the number of ways to form a committee size of n from 2n people?

Does it just count the number of ways to form a committee size of n from 2n people?

Yup!

Now … pretend the 2n people are n men and n women

Okay, so does it still mean n people regardless of gender?

Okay, so does it still mean n people regardless of gender?

Yes … the RHS is still the same …

we wouldn't muck around with that!