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Question: How many different 7-digit license plates are possible when 3 of the entries are letters and 4 are digits? Assume that repetition of the letters and numbers is allowed and that there is no restriction on where the letters or numbers can be placed.

This is how I approached the problem:

Assume I place the three letters. Then there are 4 different places I could place the numbers: _L_L_L_, represented by the underscores. The underscore can be 0,1,2, or 3 letters. This kind of problem is the "number of integer solutions of equations."

Hence, there are $\binom{n+r-1}{r-1}$ distinct nonnegative integer valued vectors $(x_1, x_2, ..., x_r)$ satisfying the equation: $x_1 + x_2 + ... + x_r = x_n \ for \ x_i \geq0$.

In this case, n=4 because you are placing 4 numbers and m=4 because there are 4 slots to place the letters. So the solution is $\binom{7}{3}*26^3*10^4$. This is in agreement with another post and matches my solution in the book: License plate consisting of 4 letters and 4 numbers

Now here is my question. Shouldn't I be able to place the numbers first and then apply the formula to the letters? I tried and am not getting the same thing. So doing the same thing: _N_N_N_N_. Here, n=3 because there are 3 letters and m=5 because there are 5 places to place. But this coefficient would give you $\binom{7}{4}$. What am I misunderstanding? Thanks in advance

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2 Answers

up vote 1 down vote accepted

${7 \choose 4}={7 \choose 3}$ and your logic is good

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Thanks for confirming my logic. –  user1527227 Apr 23 '13 at 21:11
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It turns out that ${7 \choose 3}={7\choose 4}$. More generally there is a symmetry identity that ${n \choose k}={n\choose n-k}$, for any integer $n,k$ satisfying $0\le k\le n$.

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Yeah I forgot about that. That can be seen from Pascal's Triangle because there is vertical symmetry. –  user1527227 Apr 23 '13 at 21:12
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