You are watching: How many 10 digit combinations are there

A girlfriend of mine argued why no calculate all of the possible combinations you have with $10$ number using just $4$, this method $10^4$, and also divide that for every the possible combinations you can have v $4$ digits, which method $4^4$. The results would be $39.0625$

What is wrong with the approach of the my friend"s answer?Each $256$ combinations from the 10k feasible combinations through the $10$ digits, is the results of the mix of 4 digits. If I divide $10000$ by$ 256$ shouldn"t I obtain the combinations without repeating any kind of digits?

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edited Jun 22 "13 in ~ 21:42

Sujaan Kunalan

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asked Jun 22 "13 at 21:34

Eliel van HojmanEliel van Hojman

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The native "combination" in the problem seems to suggest that the

**order**the the numbers does no matter: we have $10$ cards v the number $0,1,2, \dots,9$ written on them, and also we desire to count the number of $4$-card "hands."

Then your answer that $\dbinom104$ is perectly correct.

We point out another way of doing things that is undoubtedly acquainted to you, and also that comes close to your friend"s (incorrect) calculation.

Let us see how many ways there space to take $4$ cards from the $10$ and also lay them the end in a row. The map in the very first position have the right to be chosen in $10$ ways. For each such choice, the map in the 2nd position have the right to be favored in $9$ ways, and so on. Hence there room $$(10)(9)(8)(7)$$ways to pick $4$ cards and lay them out in a row.

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Let $H$ it is in the number of "hands" that you had counted. From your answer, we know that $H=\binom104$, but let"s pretend we don"t understand that.

For any way of the $H$ ways of picking $4$ cards, there space $4!$ means to put them the end in a row. It adheres to that $$(4!)H=(10)(9)(8)(7),$$ and also therefore$$H=\frac(10)(9)(8)(7)4!.$$

This is not much in heart from her friend"s $\frac10^44^4$. Your friend"s version likewise will not work-related if you want to count the variety of $4$-card hands wherein repetitions room allowed. The variety of strings of size $4$ is $10^4$. But hands with various numbers the repetitions carry out not all offer rise come the same number of strings of length $4$. For example, $4$ distinct cards provide us $4!$ different strings. Yet a hand with $3$ fives and $1$ eight only provides us $4$ various strings. So over there is not one **single** number that we can divide by to get the number of hands.