A Team of 3 Chosen from 5 Boys and 4 Girls — At Least One Girl

hard CBSE JEE-MAIN NCERT Class 11 Chapter 7 3 min read

Question

A team of 3 students is to be selected from a group of 5 boys and 4 girls. In how many ways can this be done if the team must have at least one girl?


Solution — Step by Step

“At least one” problems are almost always faster with the complement method. Directly counting — exactly 1 girl, exactly 2 girls, exactly 3 girls — requires three separate calculations. The complement (no girls at all) is just one.

At least one girl=Total teamsTeams with no girls\text{At least one girl} = \text{Total teams} - \text{Teams with no girls}

We have 5 + 4 = 9 students total. Order doesn’t matter (it’s a team, not a batting lineup), so we use combinations.

(93)=9!3!6!=9×8×73×2×1=84\binom{9}{3} = \frac{9!}{3! \cdot 6!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84

We’re picking 3 from the 5 boys only. Girls are completely excluded here.

(53)=5!3!2!=5×42×1=10\binom{5}{3} = \frac{5!}{3! \cdot 2!} = \frac{5 \times 4}{2 \times 1} = 10
At least one girl=8410=74\text{At least one girl} = 84 - 10 = \mathbf{74}

Why This Works

The complement principle says: everything that’s possible either contains at least one girl, or contains zero girls. These two categories cover every possible team, and they don’t overlap. So subtracting the “zero girls” count from the total gives exactly the “at least one girl” count.

This is the standard weapon for any “at least one” problem in P&C. The direct method forces you to add up cases (1 girl + 2 girls + 3 girls), which is error-prone and slow under exam conditions. The complement collapses that into a single subtraction.

The total is always (nr)\binom{n}{r} where nn is the combined group. The complement restricts nn to only the unwanted category.


Alternative Method

We can verify by direct counting — useful if you’ve already done the complement and want to cross-check.

Exactly 1 girl: (41)×(52)=4×10=40\binom{4}{1} \times \binom{5}{2} = 4 \times 10 = 40

Exactly 2 girls: (42)×(51)=6×5=30\binom{4}{2} \times \binom{5}{1} = 6 \times 5 = 30

Exactly 3 girls: (43)×(50)=4×1=4\binom{4}{3} \times \binom{5}{0} = 4 \times 1 = 4

Total=40+30+4=74\text{Total} = 40 + 30 + 4 = 74 \checkmark

Both methods agree. In the actual exam, use complement — it’s three lines versus nine.

Whenever you see “at least one” or “at least two” in a P&C problem, immediately write down: Total − Complement. This reflex alone saves 2–3 minutes per problem in JEE Main.


Common Mistake

Students often compute (93)(43)\binom{9}{3} - \binom{4}{3} thinking they’re removing “all-girl teams.” That’s wrong — (43)\binom{4}{3} is the number of teams with exactly 3 girls (all girls), not the teams with no girls at all. The complement of “at least one girl” is “zero girls,” so you subtract (53)\binom{5}{3} (choosing only from boys). Confusing the complement costs you the full 4 marks.

This exact question — or a close variant — has appeared in NCERT Class 11 Exercise 7.4 and multiple state board papers. The CBSE marking scheme specifically rewards setting up the complement equation in step 1, so write it out explicitly even if the arithmetic feels obvious.

Final Answer: 74 ways

Want to master this topic?

Read the complete guide with more examples and exam tips.

Go to full topic guide →

Try These Next