1986 AMC 12 Problem 22

Attempt Problem 22 of the 1986 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1986 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

22.

Six distinct integers are picked at random from {1,2,3,…,10}.\{1,2,3,\ldots,10\}. What is the probability that, among those selected, the second smallest is 3?3?
160\frac1{60}
16\frac16
13\frac13
12\frac12
none of these
Answer: C
Concepts:combinationprobabilityorder statistic
Small hint:
Count all six-element subsets of the ten integers
Big hint:
For second-smallest 3,3, choose one element below 33 and four above 33
Solution:
There are (106)=210\binom{10}{6}=210 possible sets. If the second-smallest element is 3,3, then 33 is selected, one element is chosen from {1,2},\{1,2\}, and four are chosen from {4,5,…,10}.\{4,5,\ldots,10\}. This gives 2(74)=70 2\binom74=70 favorable sets. The probability is 70210=13.\frac{70}{210}=\frac{1}{3}.
Thus the correct answer is C.
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