Question:
A linguistic club consists of 6 girls and 4 boys. A team of 4 members is to be selected from this group including the selection of a leader (from among these 4 members) for the team. If the team has to include at most one boy, the number of ways of selecting the team is
A linguistic club consists of 6 girls and 4 boys. A team of 4 members is to be selected from this group including the selection of a leader (from among these 4 members) for the team. If the team has to include at most one boy, the number of ways of selecting the team is
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Combinatorics Tip:
Always break "at most" or "at least" conditions into mutually exclusive exact cases.
Do not forget secondary selections! Picking the team members and picking the leader are separate, independent events that must be multiplied together.
Updated On: Apr 23, 2026
- 140
- 320
- 76
- 380
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The Correct Option is D
Solution and Explanation
Concept:
Permutations and Combinations - Selection with Conditions.
Step 1: Define the possible cases based on the given constraint. The condition states the team of 4 must include "at most one boy". This leads to two distinct cases: Case I: The team contains exactly 0 boys (which means exactly 4 girls). Case II: The team contains exactly 1 boy (which means exactly 3 girls).
Step 2: Calculate the number of ways for Case I. To select 4 girls from the available 6 girls, we use combinations: ${}^{6}C_{4} = \frac{6 \times 5}{2 \times 1} = 15$ ways. After selecting the 4 members, we must choose 1 leader from these 4: ${}^{4}C_{1} = 4$ ways. Total ways for Case I = $15 \times 4 = 60$.
Step 3: Calculate the number of ways for Case II. To select 3 girls from 6: ${}^{6}C_{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$ ways. To select 1 boy from 4: ${}^{4}C_{1} = 4$ ways. After forming the team of 4 (3 girls + 1 boy), we choose 1 leader from these 4 members: ${}^{4}C_{1} = 4$ ways. Total ways for Case II = $20 \times 4 \times 4 = 320$.
Step 4: Combine the results of both cases. Since Case I and Case II are mutually exclusive, we add their respective total ways to find the final answer.
Step 5: Calculate the final sum. Total number of ways = $60 (\text{from Case I}) + 320 (\text{from Case II}) = 380$. $$ \therefore \text{The total number of ways of selecting the team is } 380. $$
Step 1: Define the possible cases based on the given constraint. The condition states the team of 4 must include "at most one boy". This leads to two distinct cases: Case I: The team contains exactly 0 boys (which means exactly 4 girls). Case II: The team contains exactly 1 boy (which means exactly 3 girls).
Step 2: Calculate the number of ways for Case I. To select 4 girls from the available 6 girls, we use combinations: ${}^{6}C_{4} = \frac{6 \times 5}{2 \times 1} = 15$ ways. After selecting the 4 members, we must choose 1 leader from these 4: ${}^{4}C_{1} = 4$ ways. Total ways for Case I = $15 \times 4 = 60$.
Step 3: Calculate the number of ways for Case II. To select 3 girls from 6: ${}^{6}C_{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$ ways. To select 1 boy from 4: ${}^{4}C_{1} = 4$ ways. After forming the team of 4 (3 girls + 1 boy), we choose 1 leader from these 4 members: ${}^{4}C_{1} = 4$ ways. Total ways for Case II = $20 \times 4 \times 4 = 320$.
Step 4: Combine the results of both cases. Since Case I and Case II are mutually exclusive, we add their respective total ways to find the final answer.
Step 5: Calculate the final sum. Total number of ways = $60 (\text{from Case I}) + 320 (\text{from Case II}) = 380$. $$ \therefore \text{The total number of ways of selecting the team is } 380. $$
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