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The number of ways 1st, 2nd, and 3rd place prizes could be a
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05 Aug 2018, 14:51
05 Aug 2018, 14:51
Expert Reply
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Question Stats:
69% (00:45) correct
30% (00:51) wrong based on 104 sessions
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Quantity A |
Quantity B |
The number of ways 1st, 2nd, and 3rd place prizes could be awarded to 3 out of 6 contestants |
The number of ways 1st, 2nd, 3rd, 4th, and 5th place prizes could be awarded to 5 contestants |
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: The number of ways 1st, 2nd, and 3rd place prizes could be a
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11 Aug 2018, 13:45
11 Aug 2018, 13:45
Here the order of the awarded contestants matter, so we use the permutation formula:
N perm K = N!/(N-K)!
6 perm 3 = 6!/(6-3)! = 6!/3! = 6*5*4
and
5 perm 5 = 5!/(5-5)! = 5!/0! = 5*4*3*2 = 5*4*6
So they are equal. C.
N perm K = N!/(N-K)!
6 perm 3 = 6!/(6-3)! = 6!/3! = 6*5*4
and
5 perm 5 = 5!/(5-5)! = 5!/0! = 5*4*3*2 = 5*4*6
So they are equal. C.
Re: The number of ways 1st, 2nd, and 3rd place prizes could be a
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23 Aug 2018, 07:40
23 Aug 2018, 07:40
Expert Reply
Explanation
In this problem, order matters; if Mari comes in 1st place and Rohit comes in 2nd, there is a different outcome than when Rohit places 1st and Mari places 2nd. Use the fundamental counting principle to solve. To determine Quantity A, make three slots (one for each prize). Six people are available to win 1st, and then five people could win 2nd, and four people could win 3rd:
Multiply: (6)(5)(4) = 120.
For Quantity B, make 5 slots, one for each prize. Five people can win 1st prize, then 4 people can win 2nd prize, and so on:
Multiply (5)(4)(3)(2)(1) = 120. The two quantities are equal.
In this problem, order matters; if Mari comes in 1st place and Rohit comes in 2nd, there is a different outcome than when Rohit places 1st and Mari places 2nd. Use the fundamental counting principle to solve. To determine Quantity A, make three slots (one for each prize). Six people are available to win 1st, and then five people could win 2nd, and four people could win 3rd:
| 6 | 5 | 4 |
Multiply: (6)(5)(4) = 120.
For Quantity B, make 5 slots, one for each prize. Five people can win 1st prize, then 4 people can win 2nd prize, and so on:
| 5 | 4 | 3 | 2 | 1 |
Multiply (5)(4)(3)(2)(1) = 120. The two quantities are equal.
