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B
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Difficulty:
55%
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Question Stats:
60% (03:23) correct
40%
(01:37)
wrong
based on 5
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A team of 10 business analysts have to fly London tomorrow on one of two flights. 7 seats are available on the morning flight and 9 on the afternoon flight. If a list in alphabetical order is made of the names of the analysts who will fly in the morning, how many different lists are possible?
A. 330
B. 967
C. 968
D. 979
E. 1,013
A. 330
B. 967
C. 968
D. 979
E. 1,013
29 May 2026, 22:54
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The "alphabetical order" part is actually the key misdirection here. A list in alphabetical order is just a specific arrangement of whoever ends up on that flight — there's only one alphabetical ordering for any given group. So "how many different lists" is really asking how many different groups of analysts can be on the morning flight. Pure combinations.
Concept: Combinations with capacity constraints.
1. Figure out how many analysts can take the morning flight. Morning has 7 seats, afternoon has 9 seats. All 10 analysts fly, so if k go in the morning, then 10-k go in the afternoon.
2. Constraint on k: morning can hold at most 7 (k ≤ 7), afternoon can hold at most 9, so 10-k ≤ 9, which means k ≥ 1. So k ranges from 1 to 7.
3. Count the groups: We need C(10,1) + C(10,2) + C(10,3) + C(10,4) + C(10,5) + C(10,6) + C(10,7).
4. Shortcut: We know the full sum C(10,0) through C(10,10) = 2^10 = 1024. By symmetry, C(10,k) = C(10,10-k), so the sum from k=1 to 7 equals the sum from k=3 to 9. Actually the cleaner way: sum from k=0 to 10 is 1024. Subtract what we don't want.
We don't want k=0 (no one in morning), k=8 (impossible, only 7 seats), k=9, k=10.
So subtract: C(10,0) + C(10,8) + C(10,9) + C(10,10) = 1 + 45 + 10 + 1 = 57.
1024 - 57 = 967.
Answer: B
The trap most people fall into is either including k=0 (getting 1024 - 56 = 968, which is choice C) or overthinking the "alphabetical list" phrasing and multiplying by some arrangement factor. Once you realize alphabetical just fixes the ordering, it collapses to a straight combination count.
Concept: Combinations with capacity constraints.
1. Figure out how many analysts can take the morning flight. Morning has 7 seats, afternoon has 9 seats. All 10 analysts fly, so if k go in the morning, then 10-k go in the afternoon.
2. Constraint on k: morning can hold at most 7 (k ≤ 7), afternoon can hold at most 9, so 10-k ≤ 9, which means k ≥ 1. So k ranges from 1 to 7.
3. Count the groups: We need C(10,1) + C(10,2) + C(10,3) + C(10,4) + C(10,5) + C(10,6) + C(10,7).
4. Shortcut: We know the full sum C(10,0) through C(10,10) = 2^10 = 1024. By symmetry, C(10,k) = C(10,10-k), so the sum from k=1 to 7 equals the sum from k=3 to 9. Actually the cleaner way: sum from k=0 to 10 is 1024. Subtract what we don't want.
We don't want k=0 (no one in morning), k=8 (impossible, only 7 seats), k=9, k=10.
So subtract: C(10,0) + C(10,8) + C(10,9) + C(10,10) = 1 + 45 + 10 + 1 = 57.
1024 - 57 = 967.
Answer: B
The trap most people fall into is either including k=0 (getting 1024 - 56 = 968, which is choice C) or overthinking the "alphabetical list" phrasing and multiplying by some arrangement factor. Once you realize alphabetical just fixes the ordering, it collapses to a straight combination count.
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