Automobile license plates for a state consist of four letters followed by a dash and two single digits. How many different license plate combinations are possible if exactly one letter is repeated exactly once, but digits cannot be repeated? Example

We have $\binom{4}{2}$ = 6 positions for the two identical letters to occupy

And we have (25)(24) ways to choose the other two letters The total number of "words" = 6 * 26 *25*24 = 93600

Since the digits cannot be repeated = 10 * 9 = 90

The total possibilities = 93600 *90 =8,424,000

Is this correct?

  • $\begingroup$ I think it is right $\endgroup$ – Tojrah Apr 18 at 17:50

Let's say the repeated letter is $A$; you are correct that then number of ways we can put two $A$ on a plate is $6$. The are $25 \cdot 24$ ways to pick the other two letters. But what if repeated letter is $B$? We get the same number of arrangements. So I think the correct number of arranging letters is $26 \cdot 25 \cdot 24 \cdot 6$. Everything looks correct.


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