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Say that there is a 6 digit number the first digit is not allowed to be 0 or 1

so

How many number combinations start with the same two digits and end with the same three digits

ie.119333, 448222, 889888 etc..

My thoughts are 8_ 1_ 10_ 10_ 1_ 1_ = 8*10*10=800?

and how would I do it if instead of the and it said Or?

first of all, I'm I on the correct path here? any other examples similar would help.

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  • $\begingroup$ Your title should be specific to the problem at hand so that someone who has a similar question can find it if she or he searches the site. $\endgroup$ – N. F. Taussig Oct 14 '15 at 9:54
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Your count is correct.

Now let us find the number of possibilities if our number starts with the same two digits OR ends with the same three digits. Again we assume that $0$ and $1$ are forbidden as first digit. But they may occur as second digit, for example in $506888$.

There are $(8)(10^4)$ numbers that begin with two equal digits, the first (and therefore second) digit being neither $0$ or $1$.

There are $(8)(10^3)$ numbers that start with an allowed digit and whose last three digits are the same.

If we add the two numbers above, we will have double-counted the numbers in which the first two digits are the same, and the last three are the same.

It follows that the required number is $(8)(10^4)+(8)(10^3)-(8)(10^2)$.

Remark: This is a relatively simple instance of a technique called Inclusion/Exclusion. There are more elaborate versions.

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  • $\begingroup$ so Just minus the AND part? $\endgroup$ – learnmore Oct 14 '15 at 5:48
  • $\begingroup$ Yes. A general formula for two sets is $|A\cup B|=|A|+|B|-|A\cap B|$. Here $|X|$ means the number of elements in the set $X$. $\endgroup$ – André Nicolas Oct 14 '15 at 5:53
  • $\begingroup$ how would one go about in determine the count if at least one digit appears more then once in the combination.. i.e 252169 $\endgroup$ – learnmore Oct 14 '15 at 6:01
  • $\begingroup$ This is a different problem, since we no longer have two identical digits at the beginning, or three at the end. I will assume that $0$ is forbidden at the beginning. Then if we had no further restrictions, the answer would be $(9)(10^5)$. Now we subtract the bad numbers, in which all digits are distinct. Let us count the bad numbers. I will stop here and continue in the next comment. $\endgroup$ – André Nicolas Oct 14 '15 at 6:11
  • $\begingroup$ (Continued) For counting the bad numbers, there are $9$ choices for the first digit. For each of these there are $9$ choices for the second digit, since $0$ is now allowed. For each choice of first two digits, there are $8$ choices for third digit, and so on, for a total of $(9)(9)(8)(7)(6)(5)$. $\endgroup$ – André Nicolas Oct 14 '15 at 6:15

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