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01 Mar 2026, 02:47
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20 Duets: 3 Million
10 Duets: 10,000
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Team Legato is competing against Team Forte in a musical duets competition where two members from each team play a song together. Prizes are awarded for first, second, and third place to the three winning duets overall, and not for the three best duets on each individual team. Team Legato has 1/4 as many members as Team Forte.
If there are 20 duos competing, identify the closest approximate number of possible different arrangements of medal winners if it is not possible for one team to “sweep” the awards and Team Legato must come in first, and the number of possible different arrangements of medal winners if there are half as many participants
If there are 20 duos competing, identify the closest approximate number of possible different arrangements of medal winners if it is not possible for one team to “sweep” the awards and Team Legato must come in first, and the number of possible different arrangements of medal winners if there are half as many participants
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| 20 Duets | 10 Duets | |
| 10,000 | ||
| 100,000 | ||
| 1 Million | ||
| 1.5 Million | ||
| 3 Million |
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Official Answer
20 Duets: 3 Million
10 Duets: 10,000
04 Mar 2026, 23:15
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Great question to practice — this one is testing constrained permutations with complementary counting, which is exactly the kind of setup that trips people up in Two-Part Analysis because you have to nail both values independently.
Key concept being tested: Restricted arrangements using complementary counting (total minus the "forbidden" cases).
Step 1 — Understand the structure.
Team Legato has 1/4 as many members as Team Forte. With 20 duos in the competition and a 1:4 member ratio, Legato accounts for 4 duos and Forte for 16 duos. Prizes go to the top 3 duos overall.
Step 2 — Part 1 (20 duos): No sweep, Legato must come 1st.
Start with the restriction you know for certain: Legato wins 1st. That gives 4 choices (one for each Legato duo).
Now fill 2nd and 3rd from the remaining 19 duos. Use complementary counting:
- Total ways to arrange 2nd and 3rd: P(19, 2) = 19 × 18 = 342
- Subtract cases where Legato "sweeps" (2nd and 3rd also from Legato): P(3, 2) = 3 × 2 = 6
- Valid arrangements for 2nd/3rd: 342 − 6 = 336
Total for Part 1: 4 × 336 = 1,344 → closest to 1 Million.
Note: since Legato is already locked into 1st, a Forte sweep is impossible — you only need to remove the Legato sweep cases. This is where most students waste time applying a much more complicated restriction than needed.
Step 3 — Part 2 (10 duos): Half as many participants.
With 10 duos (2 Legato, 8 Forte), same setup:
- Legato wins 1st: 2 choices
- Remaining 9 duos for 2nd and 3rd. Only 1 Legato duo left — a Legato sweep of 2nd+3rd is impossible with just 1 remaining Legato duo.
- So: P(9, 2) = 9 × 8 = 72, no subtractions needed.
- Total: 2 × 72 = 144 → closest to 10,000.
Common trap: Students apply the no-sweep restriction globally — removing all cases where any one team wins all 3 medals — instead of recognizing that since Legato is already locked into 1st, only a Legato sweep remains possible. This drastically simplifies the calculation.
Takeaway: In TPA permutation problems, lock in the fixed constraint first, then apply complementary counting only to the remaining open positions.
— Kavya | 725 (99th percentile), GMAT Focus
Key concept being tested: Restricted arrangements using complementary counting (total minus the "forbidden" cases).
Step 1 — Understand the structure.
Team Legato has 1/4 as many members as Team Forte. With 20 duos in the competition and a 1:4 member ratio, Legato accounts for 4 duos and Forte for 16 duos. Prizes go to the top 3 duos overall.
Step 2 — Part 1 (20 duos): No sweep, Legato must come 1st.
Start with the restriction you know for certain: Legato wins 1st. That gives 4 choices (one for each Legato duo).
Now fill 2nd and 3rd from the remaining 19 duos. Use complementary counting:
- Total ways to arrange 2nd and 3rd: P(19, 2) = 19 × 18 = 342
- Subtract cases where Legato "sweeps" (2nd and 3rd also from Legato): P(3, 2) = 3 × 2 = 6
- Valid arrangements for 2nd/3rd: 342 − 6 = 336
Total for Part 1: 4 × 336 = 1,344 → closest to 1 Million.
Note: since Legato is already locked into 1st, a Forte sweep is impossible — you only need to remove the Legato sweep cases. This is where most students waste time applying a much more complicated restriction than needed.
Step 3 — Part 2 (10 duos): Half as many participants.
With 10 duos (2 Legato, 8 Forte), same setup:
- Legato wins 1st: 2 choices
- Remaining 9 duos for 2nd and 3rd. Only 1 Legato duo left — a Legato sweep of 2nd+3rd is impossible with just 1 remaining Legato duo.
- So: P(9, 2) = 9 × 8 = 72, no subtractions needed.
- Total: 2 × 72 = 144 → closest to 10,000.
Common trap: Students apply the no-sweep restriction globally — removing all cases where any one team wins all 3 medals — instead of recognizing that since Legato is already locked into 1st, only a Legato sweep remains possible. This drastically simplifies the calculation.
Takeaway: In TPA permutation problems, lock in the fixed constraint first, then apply complementary counting only to the remaining open positions.
— Kavya | 725 (99th percentile), GMAT Focus
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07 Mar 2026, 03:02
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I am not great at Combinatorics problems in general, but this one looked solvable on the face of it. But I am completely lost looking at the options. Must be missing some key point in understanding this problem. Would appreciate any help here.
