Combinatorics interview questions

We have 8 points on a circle, so there are $\binom{8}{2}=28$ chords in total. Choosing 4 chords uniformly at random gives $\binom{28}{4}$ equally likely outcomes.

For the 4 chords to form a convex quadrilateral, their union must be exactly the boundary of that quadrilateral. Because the points lie on a circle, any 4 distinct points determine a convex quadrilateral whose sides are the chords connecting consecutive points in cyclic order. Thus a favorable set of 4 chords corresponds precisely to choosing 4 of the 8 points and taking the four boundary chords of that set. There are $\binom{8}{4}=70$ ways to choose the points, and each yields exactly one favorable 4‑chord set. No other set of 4 chords can form a convex quadrilateral, since any quadrilateral must have four distinct vertices, and on a circle the only non‑self‑intersecting 4‑cycle on four points is the boundary cycle.

Hence the probability is

We have 15 boys and 10 girls, totaling 25 students. The students line up uniformly at random. We want the expected number of adjacent positions where one is a boy and the other is a girl (order BG or GB).

Let $X$ be the number of adjacent boy-girl pairs. For each adjacent pair of positions $(i, i+1)$ with $i = 1,\dots,24$, define an indicator $I_i$ that the two students at those positions are of opposite sexes. Then $X = \sum_{i=1}^{24} I_i$, and by linearity of expectation,

By symmetry, the probability is the same for every adjacent pair; denote it by $p$. Hence $\mathbb{E}[X] = 24 \cdot p$.

Now compute $p$. There are $25 \times 24$ equally likely ordered assignments of two distinct students to the two positions. Favorable cases are (boy, girl) and (girl, boy). There are $15 \times 10 = 150$ ways to choose a boy then a girl, and $10 \times 15 = 150$ ways to choose a girl then a boy, for a total of $300$ favorable ordered pairs. Thus

Alternatively, $\Pr(\text{boy then girl}) = \frac{15}{25} \cdot \frac{10}{24} = \frac{1}{4}$, and $\Pr(\text{girl then boy}) = \frac{10}{25} \cdot \frac{15}{24} = \frac{1}{4}$, so $p = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$.

Therefore,

Let $M$, $T$, $W$ be the events that Monday, Tuesday, Wednesday produce a matching pair, respectively. We want

1. $P(M^c)$. Total ways to choose $2$ socks from $10$: $\binom{10}{2}=45$. Matching pairs: $5$ (one per color). So $P(M)=5/45=1/9$, hence

2. $P(T^c \mid M^c)$. After a non‑matching Monday, two socks of different colors are removed. The remaining $8$ socks consist of $3$ colors with $2$ socks each and $2$ colors with $1$ sock each. Total pairs from $8$: $\binom{8}{2}=28$. Only the three colors with $2$ socks can produce a matching pair, so $3$ matching pairs. Thus $P(T\mid M^c)=3/28$ and

3. $P(W \mid M^c \cap T^c)$. After Monday and Tuesday are both non‑matching, the composition of the remaining $6$ socks depends on how Tuesday's non‑matching pair was formed. Starting from the Monday state ($3$ colors with $2$ socks, $2$ colors with $1$ sock), Tuesday's non‑matching pair can be of three types:

• Case A: Both socks from the $2$-sock colors. Number of ways: $\binom{3}{2}\cdot2\cdot2 = 12$. Resulting state: $1$ color with $2$ socks, $4$ colors with $1$ sock. Probability of a match on Wednesday: $\frac{1}{\binom{6}{2}} = \frac{1}{15}$.
• Case B: One sock from a $2$-sock color and one from a $1$-sock color. Number of ways: $3\cdot2\cdot2\cdot1 = 12$. Resulting state: $2$ colors with $2$ socks, $2$ colors with $1$ sock, $1$ color with $0$ socks. Probability of a match on Wednesday: $\frac{2}{15}$.
• Case C: Both socks from the two $1$-sock colors. Number of ways: $1$. Resulting state: $3$ colors with $2$ socks, $2$ colors with $0$ socks. Probability of a match on Wednesday: $\frac{3}{15} = \frac15$.

Total non‑matching pairs on Tuesday: $12+12+1 = 25$, so the conditional probabilities of the cases given $T^c$ are $\frac{12}{25},\frac{12}{25},\frac{1}{25}$. Hence

4. Multiply.

Thus the desired probability is $\boxed{\frac{26}{315}}$.

(One can verify by counting all sequences: total sequences $\binom{10}{2}\binom{8}{2}\binom{6}{2}=18900$, favorable sequences $5\cdot24\cdot13=1560$, giving the same fraction.)