Maths Olympiad Prep

Track / Stage 5 / 82 of 400 #682 of 1964

Problem 682

AIME late
Combinatorics Difficulty 5.2 Find the answer

8. Six boys and aa girls stand in a row at random, with each boy being adjacent to at least one other boy, and the probability that at least four boys stand together is pp. If p1100p \leqslant \frac{1}{100}, then the minimum value of aa is . \qquad

A number or a short expression. Spacing and $ signs are ignored.

Official solution

8.594.

Notice that,
p=2Ca+12+(a+1)Ca+13+3Ca+12+(a+1)=6a+6a2+8a+61100. \begin{aligned} p & =\frac{2 \mathrm{C}_{a+1}^{2}+(a+1)}{\mathrm{C}_{a+1}^{3}+3 \mathrm{C}_{a+1}^{2}+(a+1)} \\ & =\frac{6 a+6}{a^{2}+8 a+6} \leqslant \frac{1}{100} . \end{aligned}

Therefore, f(a)=a2592a5940f(a)=a^{2}-592 a-594 \geqslant 0.
Also, f(0)0f(0)0, thus, the minimum value of aa is 594.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.