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12 Apr 2021, 06:15
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C
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Difficulty:
15%
(low)
Question Stats:
83% (01:51) correct
17%
(02:09)
wrong
based on 196
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A badminton club has 21 members. What is the ratio of number of 6-member committees that can be formed from the members of the club to the number of 5-member committees that can be formed from the members of the club?
(A) 17 to 16
(B) 15 to 1
(C) 8 to 3
(D) 17 to 6
(E) 16 to 5
(A) 17 to 16
(B) 15 to 1
(C) 8 to 3
(D) 17 to 6
(E) 16 to 5
12 Apr 2021, 12:41
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Ratio of possible combinations of 6-member comittees to 5-member comittees:
\(\frac{21C6}{21C5}\)
\(\frac{21!}{6!*15!}/\frac{21!}{5!*16!}\), simplifies to \(\frac{16}{6} = \frac{8}{3}\)
Imo C - I would appreciate it if you corrected me in case I am wrong.
\(\frac{21C6}{21C5}\)
\(\frac{21!}{6!*15!}/\frac{21!}{5!*16!}\), simplifies to \(\frac{16}{6} = \frac{8}{3}\)
Imo C - I would appreciate it if you corrected me in case I am wrong.
03 Jun 2022, 02:50
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My 2 cents on this:
Total members: 21
Creating 6 members committee = 21C6
Creating 5 members committee 21C5
So, Divide 21C6 / 21C5; what we get is a bunch of numbers getting eliminated in the process. Remaining fraction is 16/6 which is simplified to 8/3.
Hence C
Total members: 21
Creating 6 members committee = 21C6
Creating 5 members committee 21C5
So, Divide 21C6 / 21C5; what we get is a bunch of numbers getting eliminated in the process. Remaining fraction is 16/6 which is simplified to 8/3.
Hence C
