What is the combinatoric explain of the equation $4{51 \choose 12}={52 \choose 13}$

Question

I don't know how to explain the combinatoric meaning of the expression $4{51 \choose 12} = {52 \choose 13}$. I want a explain using cards desk? Thank you very much.

Answer 1

So suppose you want to choose $13$ cards out of $52$, this is normally done in $\binom{52}{13}$ ways.

Then, you fix one card , which is done in $52$ ways, and then choose $12$ cards out of the remaining $51$. This can be done in $52\binom{51}{12}$ ways. But, we have to divide by $13$, and the reason is this: in the end, we end in a collection of cards. Call these cards from card $1$ to card $13$.

See that, if I had picked card $1$ first, and then while picking $12$ cards, picked up the cards $2,3,4,...,13$, then my final collection is $1,2,...,13$.

But, it's also possible that I picked card $2$ first, and then while picking $12$ cards, picked up the cards $1,3,4,...,13$, then my final collection is again $1,2,...,13$.

But, it's also possible that I picked card $8$ first, and then while picking $12$ cards, picked up the cards $1,3,4,5,6,7,9,10,11,12,13$, then my final collection is again $1,2,...,13$.

Do you see what I am hinting at? The same collection is generated $13$ times using our previous algorithm, because each one of the elements in the collection could have been the first card. Hence, we have to divide by $13$, because all these cases are resulting in the same collection.

Hence, $\binom{52}{13} = 52\binom{51}{12} \div 13 = 4\binom{51}{12}$.

I hope this was clear.

Answer 2

It means there is four times more ways to pick $13$ cards from a deck of $52$, than to pick $12$ cards from a deck of $51$.

Answer 3

If you are dealt $13$ cards the probability that the ace of hearts is among them is ${1\over4}\>$.

Answer 4

You may interpret it as $4 = \frac{52}{13}$, and use the identity

$$k\binom{n}{k} = n \binom{n-1}{k-1}$$

Which can be explained in that way : the left-hand-side is counting the ways to pick $k$ elements out of $n$ and chosing a special element among them. The right-hand-side is picking the special element first, and the $k-1$ others afterwards, among the $n-1$ remaining.

Answer 5

You can have $\binom{52}{13}$ possible hands.
Divide the cards into lots of $4$. In each round, you could get any of the $4$.

Imagine that you are dealt last, and that $12$ cards each have been dealt. When they finally come to you, $51$ cards have been dealt, but you could get any of the last 4 to complete your hand.

Thus $4\binom{51}{12} = \binom{52}{13}$