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An inventory of coins contains 100 different coins.
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05 Aug 2018, 14:53
05 Aug 2018, 14:53
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Expert Reply
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Question Stats:
69% (00:37) correct
30% (00:53) wrong based on 140 sessions
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An inventory of coins contains 100 different coins.
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
The number of possible collections of 56 coins that can be selected (the order of the coins does not matter) |
The number of possible collections of 44 coins that can be selected (the order of the coins does not matter) |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Re: An inventory of coins contains 100 different coins.
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11 Aug 2018, 13:42
11 Aug 2018, 13:42
1
Combinations! Order doesn't matter.
As long as you remember the formula:
N choose K = N!/(K!(N-K)!)
You will quickly solve questions like this.
Option A:
100 choose 56 = 100!(56!(100-56)!) = 100!/(56!44!)
Option B:
100 choose 44 = 100!(44!(100-44)!) = 100!/(44!56!)
So they are equal. Don't try to solve the factorial, as the number of combinations are astronomical.
C.
As long as you remember the formula:
N choose K = N!/(K!(N-K)!)
You will quickly solve questions like this.
Option A:
100 choose 56 = 100!(56!(100-56)!) = 100!/(56!44!)
Option B:
100 choose 44 = 100!(44!(100-44)!) = 100!/(44!56!)
So they are equal. Don't try to solve the factorial, as the number of combinations are astronomical.
C.
Re: An inventory of coins contains 100 different coins.
[#permalink]
23 Aug 2018, 07:42
23 Aug 2018, 07:42
3
Expert Reply
Explanation
This is a classic combinatorics problem in which order doesn’t matter—in fact, the problem states that explicitly. Use the standard “order doesn’t matter” formula:
\(\frac{total!}{in! \times out!}\)
For Quantity A:
\(\frac{100!}{56! \times 44!}\)
Because the numbers are so large, there must be a way to solve the problem without actually simplifying (even with a calculator, this is unreasonable under GRE time limits). Try Quantity B and compare:
\(\frac{100!}{44! \times 56!}\)
The quantities are equal. Note that this will always work—when order doesn’t matter, the number of ways to pick 56 and leave out 44 is the same as the number of ways to pick 44 and leave out 56. Either way, it’s one group of 56 and one group of 44. What actually happens to those groups (being part of a collection, being left out of the collection, etc.) is irrelevant to the ultimate solution.
This is a classic combinatorics problem in which order doesn’t matter—in fact, the problem states that explicitly. Use the standard “order doesn’t matter” formula:
\(\frac{total!}{in! \times out!}\)
For Quantity A:
\(\frac{100!}{56! \times 44!}\)
Because the numbers are so large, there must be a way to solve the problem without actually simplifying (even with a calculator, this is unreasonable under GRE time limits). Try Quantity B and compare:
\(\frac{100!}{44! \times 56!}\)
The quantities are equal. Note that this will always work—when order doesn’t matter, the number of ways to pick 56 and leave out 44 is the same as the number of ways to pick 44 and leave out 56. Either way, it’s one group of 56 and one group of 44. What actually happens to those groups (being part of a collection, being left out of the collection, etc.) is irrelevant to the ultimate solution.
