Selections with an 'At Least One' Condition

“At least one” is the signal for the complement method: instead of adding up every qualifying case, count everything and subtract the one case you do not want. The complement of “at least one woman” is “no women at all”, a single easy count.

A team of $4$ is to be selected from $6$ men and $4$ women. How many possible teams contain at least one woman? [4]

The working

Step 1, count all possible teams of $4$ from the $10$ people: $^{10}C_{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210 \quad \text{(M1)}$

Step 2, count the unwanted case, all men (no women), i.e. $4$ from the $6$ men: $^{6}C_{4} = \frac{6 \times 5 \times 4 \times 3}{4 \times 3 \times 2 \times 1} = 15 \quad \text{(M1)}$

Step 3, subtract to leave the teams with at least one woman: $210 - 15 = 195 \quad \text{(M1, A1)}$

Where the marks are won and lost

• Recognising the complement (“at least one” $=$ total $-$ none) is the efficient route. Adding the $1$-, $2$-, $3$- and $4$-woman cases separately is valid but far longer and more error-prone.
• The “none” case is all men: $^{6}C_{4}$, choosing the whole team from the $6$ men only.
• The total is $^{10}C_{4}$ over all ten people, do not restrict it prematurely.

Common mistakes

• Computing “exactly one woman” and stopping, that is not “at least one”.
• Using $^{4}C_{4}$ (all women) as the unwanted case instead of all men.
• Adding when the complement method calls for subtraction.

Full method: Arrangements & Selections notes. See also Combinations (nCr). Topic home: Permutations & Combinations pillar.