Permutations and Combinations: Question 4

Worked solution

Part (a): No restriction

A team of $6$ is selected from $12$ researchers, and the order in which they are chosen doesn’t matter. This is a combination:

$\binom{12}{6} = \frac{12!}{6!\,6!} = 924.$

Check (Pascal-style build-up): $\binom{12}{6}=\binom{11}{5}+\binom{11}{6}=462+462=924$, consistent with the standard value of $\binom{12}{6}$.

Part (b): Exactly 2 junior researchers (and 4 senior researchers)

Choose $2$ of the $5$ junior researchers, and independently choose $4$ of the $7$ senior researchers. By the multiplication principle:

$\binom{5}{2}\times\binom{7}{4} = 10\times35 = 350.$

Check: $\binom{5}{2}=\dfrac{5\times4}{2\times1}=10$ and $\binom{7}{4}=\binom{7}{3}=\dfrac{7\times6\times5}{3\times2\times1}=35$, so $10\times35=350$, confirming the result.

Part (c): At least 4 senior researchers (casework)

Since the team has exactly $6$ people chosen from $7$ seniors and $5$ juniors, “at least $4$ senior” splits into three cases, according to how many seniors are on the team:

Case 1: $4$ senior, $2$ junior $\binom{7}{4}\times\binom{5}{2} = 35\times10 = 350$

Case 2: $5$ senior, $1$ junior $\binom{7}{5}\times\binom{5}{1} = 21\times5 = 105$

Case 3: $6$ senior, $0$ junior $\binom{7}{6}\times\binom{5}{0} = 7\times1 = 7$

Adding the cases (they can’t overlap, since a team has one definite number of seniors): $350+105+7 = 462.$

Check (complement): the complement of “at least $4$ senior” is “at most $3$ senior”, i.e. $1$, $2$ or $3$ seniors (senior $=0$ is impossible here, since that would need $6$ juniors but only $5$ exist):

• $1$ senior, $5$ junior: $\binom{7}{1}\times\binom{5}{5} = 7\times1 = 7$
• $2$ senior, $4$ junior: $\binom{7}{2}\times\binom{5}{4} = 21\times5 = 105$
• $3$ senior, $3$ junior: $\binom{7}{3}\times\binom{5}{3} = 35\times10 = 350$

Complement total: $7+105+350 = 462$. Since every team has either “at least $4$ senior” or “at most $3$ senior” (and not both), these two totals should add to the grand total from part (a): $462+462 = 924 = \binom{12}{6}.$ This matches part (a) exactly, confirming $462$.

Final answers

• (a) No restriction $= \boxed{924}$
• (b) Exactly $2$ junior (and $4$ senior) $= \boxed{350}$
• (c) At least $4$ senior $= \boxed{462}$