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Suppose we have a set of 12 alphabets. And we are required to form a 7 letter word using the 12 alphabets. Now, since repetition is allowed, the total no. of arrangements would be $\ 12^7$. The total no. of arrangements without any repetition would be $\ P_7^{12}$. From these two results, we can deduce the no. of arrangements with at least one repetition to be $\ 12^7$ - $\ P_7^{12}$. But how can we find something specific like, the total no. of arrangements such that 4 and only 4 of the letters are same (7 letter word).

x80W
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  • One way to go about it would be subtracting the probability of that event not happening (this will be a sum of several others happening) from the total. – Elen Khachatryan Oct 10 '19 at 05:44

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There are 12 ways to select a letter that will be repeated exactly 4 times. We are left with 11 alphabets from which we have to choose 3 distinct letters. This can be done in $\binom{11}{3}$ ways.

Once the letters are selected, fix the fours positions of the letter that repeats. This can be done in $\binom{7}{4}$ ways. The remaining 3 distinct letters can be permuted in $3! = 6$ ways to fill the remaining three spots.

Our final answer is $12 \cdot 6 \cdot \binom{11}{3} \cdot \binom{7}{4}$.

mrpandey
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  • This actually exceeds the total no. of permutations possible (repetition allowed). – x80W Oct 11 '19 at 02:18
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    @x80W My answer evaluates to $415800$ which is definitely less than total no. of permutations possible with repetition allowed ($12^7 = 35831808$). – mrpandey Oct 11 '19 at 11:16