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The problem is,

Let $S$ be the set of five-digit numbers formed by the digits $1, 2, 3, 4$ and $5$, using each digit exactly once such that exactly two odd positions are occupied by odd digits. What is the sum of the digits in the rightmost position of the numbers in S?

I am looking for some approach for solving this problem.

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    Suppose $X_1X_2X_3X_4X_5$ is the 5-digit number. Then if we pick the number at random from $S$, then $X_1X_3X_5$ is uniformly distributed in $132, 134, 352, 354, 512, 514$ and all their permutations (e.g.: $132, 123, 213, 231, 312, 321$, and so on). Then, $E[X_1+X_3+X_5] = \frac{1}{6}(6+8+10+12+8+10) = 9$. Finally, note that the distributions of $X_1$, $X_3$, and $X_5$ are identical; so $E[X_5]=3$. Hence, the required answer is $3|S|$.2011-08-13

2 Answers 2