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25 Jul 2024, 01:30
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
75% (01:49) correct
25%
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Alan has a flock of sheep from which he will choose 4 to take with him to the livestock show in Houston. If Alan has 15 distinct possible groups of sheep he could take to the show, then which of the following is the number of sheep in his flock?
A. 30
B. 15
C. 7
D. 6
E. 5
A. 30
B. 15
C. 7
D. 6
E. 5
25 Jul 2024, 01:58
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I think D.
Lets assume he has n number of sheeps, he will choose 4 to take with him to the livestock show in Houston.
Total possibilities of selecting 4 out of n = nC4 = 15
now put values from options and check you will notice Option D satisfies the given condition.
\(6C4 = \frac{6!}{2!4!} = 15\)
Answer is option D.
Lets assume he has n number of sheeps, he will choose 4 to take with him to the livestock show in Houston.
Total possibilities of selecting 4 out of n = nC4 = 15
now put values from options and check you will notice Option D satisfies the given condition.
\(6C4 = \frac{6!}{2!4!} = 15\)
Answer is option D.
25 Jul 2024, 02:08
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Let \(x\) be the number of sheep.
Choosing 4 sheep from \(x\) sheep will have 15 distinct possible groups of sheep: \(\frac{x!}{4!(x-4)!} = 15\)
Looking at the answer choices, one can eliminate A and B immediately as it is impossible for the value of \(x\) to be equal to or greater than 15 given that the answer of the equation is 15. Plugging in 7 (Choice C) for \(x\) will give \(\frac{7!}{3!}\) which, without doing any working out, one can see will be greater than 15.
This leaves D & E. One can eliminate E, as when \(x = 5\) the equation will yield an answer of 5. Therefore the answer will have to be D.
Plugging in 6 (Choice D): \(\frac{6!}{4!(2)!} = 15\)
\(\frac{6*5}{2} = 15\)
\(15 = 15\)
ANSWER D
Choosing 4 sheep from \(x\) sheep will have 15 distinct possible groups of sheep: \(\frac{x!}{4!(x-4)!} = 15\)
Looking at the answer choices, one can eliminate A and B immediately as it is impossible for the value of \(x\) to be equal to or greater than 15 given that the answer of the equation is 15. Plugging in 7 (Choice C) for \(x\) will give \(\frac{7!}{3!}\) which, without doing any working out, one can see will be greater than 15.
This leaves D & E. One can eliminate E, as when \(x = 5\) the equation will yield an answer of 5. Therefore the answer will have to be D.
Plugging in 6 (Choice D): \(\frac{6!}{4!(2)!} = 15\)
\(\frac{6*5}{2} = 15\)
\(15 = 15\)
ANSWER D
