# Combinations with 10 digits

To find the number of $4$ digit combinations I can form with $10$ digits, without repeating any of the digits, I run the binomial coefficient of $10$ and $4$ and get $210$.

A friend of mine suggested why not calculate all of the possible combinations you have with $10$ digits using only $4$, this means $10^4$, and divide it for all the possible combinations you can have with $4$ digits, which means $4^4$. The results would be $39.0625$

What is wrong with the approach of the my friend's answer? Each $256$ combinations from the 10k possible combinations with the $10$ digits, is the results of the combination of 4 digits. If I divide $10000$ by$256$ shouldn't I get the combinations without repeating any digits?

• One immediate reason that your friend's approach can't possibly be right is that it doesn't offer a whole number! You know that the answer has to be a whole number, so any approach that yields a fraction like $39.0625$ must be 'counting' incorrectly somewhere along the way. Commented Jun 22, 2013 at 22:55

The word "combination" in the problem seems to indicate that the order of the numbers does not matter: we have $10$ cards with the digits $0,1,2, \dots,9$ written on them, and we want to count the number of $4$-card "hands."

Then your answer of $\dbinom{10}{4}$ is perectly correct.

We mention another way of doing things that is undoubtedly familiar to you, and that comes close to your friend's (incorrect) calculation.

Let us see how many ways there are to take $4$ cards from the $10$ and lay them out in a row. The card in the first position can be chosen in $10$ ways. For each such choice, the card in the second position can be chosen in $9$ ways, and so on. Thus there are $$(10)(9)(8)(7)$$ ways to choose $4$ cards and lay them out in a row.

Let $H$ be the number of "hands" that you had counted. From your answer, we know that $H=\binom{10}{4}$, but let's pretend we don't know that.

For any way of the $H$ ways of choosing $4$ cards, there are $4!$ ways to lay them out in a row. It follows that $$(4!)H=(10)(9)(8)(7),$$ and therefore $$H=\frac{(10)(9)(8)(7)}{4!}.$$

This is not far in spirit from your friend's $\frac{10^4}{4^4}$. Your friend's version also will not work if you want to count the number of $4$-card hands where repetitions are allowed. The number of strings of length $4$ is $10^4$. But hands with different numbers of repetitions do not all give rise to the same number of strings of length $4$. For example, $4$ distinct cards give us $4!$ different strings. But a hand with $3$ fives and $1$ eight only gives us $4$ different strings. So there is not one single number that we can divide by to get the number of hands.

• Thanks! I see now that the error was to consider that four digits will always have 4^4 different combinations, but following the example you gave, given four same digits we will only have one possible combination. Commented Jun 23, 2013 at 7:33

Your friend is getting rid of options which should not be left out. $4^4$ does not only give you options in which a certain number appears more than once. For instance, if I were to select from the numbers $1$ to $4$ for four times, then $4^4$ would also include the option $1 2 3 4$. Hence, he is also deviding out options which he shouldn't leave out. It's better to stick to the definition as you did.