Probability

Given distinct real values $a < b < c < d < e$ , how many orderings $x_1,x_2,x_3,x_4,x_5$ of these values are there such that $x_1 > x_2 > x_3 < x_4 < x_5$ ? First note that we must have $x_3 = a$ , as the inequalities imply $x_3$ is smaller than each of the other $4$ values. Then there are $\binom{4}{2} = 6$ ways to select $x_1,x_2$ from $\{b,c,d,e\}$ (note that for any two values we pick, the ordering is specified because they are all distinct values); from this selection $x_4,x_5$ are implied as well. This shows that for $a < b < c < d < e$ , there are 6 ways to order the values to satisfy the desired inequality chain.

Returning to the problem at hand, there are $100 \times 99 \times 98 \times 97 \times 96$ unique selections of coupons. Note that there are $\binom{100}{5}$ possible combinations of coupon selections, and for each of these combinations we have from the above that there are $6$ valid orderings. Thus the probability of such an ordering occurring is $\frac{6\times \binom{100}{5}}{100 \times 99 \times 98 \times 97 \times 96} = \frac{1}{20}$ , giving us the solution of $21$ .

One way to see this problem is to take 5 numbers at random from the set { $1,2,3,...,100$ } and assign them to $x_1, x_2, ... , x_5$ such that the given constraints are satisfied. $x_3$ is the minimum so its position is fixed while there is an ordering for $x_1, x_2$ and $x_4, x_5$ . So the largest number can be either $x_1$ or $x_5$ . For the case when $x_1$ is largest, choosing a place for $x_2$ fixes the places for $x_4, x_5$ . So there are $3$ possibilities in this case. Similarly for the case when $x_5$ is largest, we have $3$ possibilities. Total number of permutations of $x_1, ... , x_5$ is simply $5! = 120$ . Thus we obtain the required probability as $\frac{6}{120} = \frac{1}{20}$