Permutations and Combinations: Question 5

Worked solution

Step 1: Choose the counting method

“At least one girl” is awkward to count directly, since it covers the cases of $1$, $2$, $3$ or $4$ girls on the delegation. It’s much quicker to use the complement:

$\text{(at least one girl)} = \text{(all delegations)} - \text{(no girls at all)}.$

Step 2: Count the total number of delegations

Choosing $4$ members from all $10$, with no restriction:

$\binom{10}{4} = \frac{10!}{4!\,6!} = 210.$

Step 3: Count the delegations with no girls (all boys)

Choosing $4$ members from the $6$ boys, with none of the $4$ girls included:

$\binom{6}{4} = \binom{6}{2} = \frac{6\times5}{2\times1} = 15.$

Step 4: Subtract

$210 - 15 = 195.$

Step 5: Check by direct casework

Splitting by the number of girls on the delegation and adding the four possible cases:

• $1$ girl, $3$ boys: $\binom{4}{1}\times\binom{6}{3} = 4\times20 = 80$
• $2$ girls, $2$ boys: $\binom{4}{2}\times\binom{6}{2} = 6\times15 = 90$
• $3$ girls, $1$ boy: $\binom{4}{3}\times\binom{6}{1} = 4\times6 = 24$
• $4$ girls, $0$ boys: $\binom{4}{4}\times\binom{6}{0} = 1\times1 = 1$

$80+90+24+1 = 195.$

Both methods agree: $195$.

Why the other options are wrong

• A ($15$): this is $\binom{6}{4}$, the number of all-boy delegations, the complement itself, not the answer to “at least one girl”.
• C ($210$): this is $\binom{10}{4}$, the total number of delegations with no restriction at all. The “at least one girl” condition was never applied.
• D ($186$): this comes from miscalculating $\binom{6}{4}$ as $6\times4=24$ instead of $15$, giving $210-24=186$.

Final answer

• Number of delegations with at least one girl $= \boxed{195}$, option B.