(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Two n-digit integers (leading zeros allowed) are considered equivalent if one is a rearrangement of the other. (For example, 12033, 20331, and 01332 are considered equivalent five-digit integers.) If the digits 1, 3, and 7 can appear at most once, how many

nonequivalent five-digit integers are there?

(Problem 11 from Chapter 1.4 of Grimaldi's DISCRETE AND COMBINATORIAL MATHEMATICS)

Noted for any future use it might serve.

2. Relevant equations

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3. The attempt at a solution

I know that effectively only 7 integers are being considered. That would give a base number to our solution of C(11, 5) where 7 (before the n + r -1 procedure in the combination formula) is the number of distinct items in our combination and 5 the number of repetitions. What I don't know is what to do with the remaining 3 integers. I know the answer from looking at the back of the book, but can't reason how it was derived.

The answer (forgive me if my notation isn't correct):

C(11, 5) + 3*C(10, 4) + 3*C(9, 3) + 3*C(8, 2)

I guess the further additions after the initial C(11, 5) account for them, but I don't have the faintest idea why that is so. Thanks.

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# Homework Help: Combinatorics Two n-digit integers Question

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