Permutations and Combinations: Question 7

Worked solution

Part (a): No restriction

A team of $5$ is selected from $12$ students, and the order of selection doesn’t matter. This is a combination:

$\binom{12}{5} = \frac{12!}{5!\,7!} = 792.$

Part (b): Both Farah and Idris included

If Farah and Idris are both already on the team, that uses up $2$ of the $5$ places, leaving $3$ places to be filled from the other $10$ students:

$\binom{10}{3} = \frac{10\times9\times8}{3\times2\times1} = 120.$

Check: $\binom{10}{3}=\binom{10}{7}$, and building up from Pascal’s triangle-style values, $\binom{10}{3}=120$ is the standard value, confirming the result.

Part (c): Farah and Idris not both included (complement)

The teams in which Farah and Idris are not both included are exactly the total teams minus the teams found in part (b) where they are both included:

$792 - 120 = 672.$

Check (direct casework): split by how many of Farah and Idris are on the team. Either neither is included, or exactly one is included (both cases satisfy “not both together”):

• Neither Farah nor Idris: choose all $5$ team members from the other $10$ students: $\binom{10}{5}=252$.
• Exactly one of Farah/Idris: there are $2$ choices for which one of them is on the team (Farah-not-Idris, or Idris-not-Farah), and for each, the remaining $4$ places come from the other $10$ students: $2\times\binom{10}{4}=2\times210=420$.

$252+420 = 672.$

Both methods agree: $672$.

Final answers

• (a) No restriction $= \boxed{792}$
• (b) Both Farah and Idris included $= \boxed{120}$
• (c) Not both included $= \boxed{672}$