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Ther are 8 different flowers for sale at the store
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25 Nov 2019, 08:14
25 Nov 2019, 08:14
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65% (00:53) correct
34% (00:52) wrong based on 52 sessions
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Ther are 8 different flowers for sale at the store.
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation
Quantity A |
Quantity B |
Twice the total possible bouquets of 2 different flowers |
The total possible bouquets of 5 different flowers |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation
Re: Ther are 8 different flowers for sale at the store
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25 Nov 2019, 08:46
25 Nov 2019, 08:46
1
8C2 = 28, so twice that is 56 =A
8C5 = 56 = B
Answer C, they're equal.
8C5 = 56 = B
Answer C, they're equal.
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Ther are 8 different flowers for sale at the store
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18 Jan 2024, 02:08
18 Jan 2024, 02:08
1
The number of total possible bouquets is a combination because the order of flowers doesn't matter.
The combination formula is \(\frac{n! }{ (n-r)! r!}\), where \(n\) is the total number of options (in this case, the number of different flowers) and \(r\) is the number of options being picked (number of flowers in each bouquet).
The total possible bouquets of 2 different flowers is:
\(\frac{8! }{ (8-2)! 2!} \)
\(= \frac{8! }{ 6! 2!} \)
\(= \frac{8*7*6*5*4*3*2*1}{ (6*5*4*3*2*1)(2*1)} \)
\(= \frac{8*7}{2}\)
\(= 28\)
Quantity A is twice that, so it's 56.
The total possible bouquets of 5 different flowers is:
\(\frac{8! }{ (8-5)! 5!} \)
\(= \frac{8! }{ 3! 5!} \)
\(= \frac{8*7*6*5*4*3*2*1}{ (5*4*3*2*1)(3*2*1)} \)
\(= \frac{8*7*6}{6}\)
\(= 56\)
Quantity B is 56.
Thus, Quantity A = Quantity B.
The combination formula is \(\frac{n! }{ (n-r)! r!}\), where \(n\) is the total number of options (in this case, the number of different flowers) and \(r\) is the number of options being picked (number of flowers in each bouquet).
The total possible bouquets of 2 different flowers is:
\(\frac{8! }{ (8-2)! 2!} \)
\(= \frac{8! }{ 6! 2!} \)
\(= \frac{8*7*6*5*4*3*2*1}{ (6*5*4*3*2*1)(2*1)} \)
\(= \frac{8*7}{2}\)
\(= 28\)
Quantity A is twice that, so it's 56.
The total possible bouquets of 5 different flowers is:
\(\frac{8! }{ (8-5)! 5!} \)
\(= \frac{8! }{ 3! 5!} \)
\(= \frac{8*7*6*5*4*3*2*1}{ (5*4*3*2*1)(3*2*1)} \)
\(= \frac{8*7*6}{6}\)
\(= 56\)
Quantity B is 56.
Thus, Quantity A = Quantity B.
