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goodyear2013
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If a 3-member subcommittee is to be formed from a certain 6- Updated on: 01 Apr 2014, 04:57
Originally posted by goodyear2013 on 01 Apr 2014, 03:37.
Last edited by Bunuel on 01 Apr 2014, 04:57, edited 1 time in total.
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C
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If a 3-member subcommittee is to be formed from a certain 6-member committee, how many different such subcommittee are possible?

A. 6
B. 18
C. 20
D. 108
E. 216
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C
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If a 3-member subcommittee is to be formed from a certain 6- 01 Apr 2014, 12:16
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goodyear2013
If a 3-member subcommittee is to be formed from a certain 6-member committee, how many different such subcommittee are possible?

A. 6
B. 18
C. 20
D. 108
E. 216
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Another way:
1st member can be selected in 6 ways
2nd can be selected in 5 ways
3rd can be selected in 4 ways
So total ways : 120
But to avoid the similar scenarios 120/3!=20
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Bunuel
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If a 3-member subcommittee is to be formed from a certain 6- 01 Apr 2014, 04:56
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goodyear2013
if a 3-member subcommittee is to be formed from a certain 6-member committee, how many different such subcommittee are possible?

A) 6
B) 18
C) 20
D) 108
E) 216
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Choosing 3 out of 6: \(C^3_6=\frac{6!}{(6-3)!3!}=20\).

Answer: C.
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If a 3-member subcommittee is to be formed from a certain 6- 22 Mar 2017, 07:40
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goodyear2013
If a 3-member subcommittee is to be formed from a certain 6-member committee, how many different such subcommittee are possible?

A. 6
B. 18
C. 20
D. 108
E. 216
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\(6C_3 = 20\)

Thus, the answer must be (C) 20
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If a 3-member subcommittee is to be formed from a certain 6- 01 May 2018, 10:00
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goodyear2013
If a 3-member subcommittee is to be formed from a certain 6-member committee, how many different such subcommittee are possible?

A. 6
B. 18
C. 20
D. 108
E. 216
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The number of possible subcommittees is 6C3 = (6 x 5 x 4)/3! = 20.

Answer: C
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If a 3-member subcommittee is to be formed from a certain 6- 25 Aug 2018, 05:43
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Bunuel
goodyear2013
if a 3-member subcommittee is to be formed from a certain 6-member committee, how many different such subcommittee are possible?

A) 6
B) 18
C) 20
D) 108
E) 216

Choosing 3 out of 6: \(C^3_6=\frac{6!}{(6-3)!3!}=20\).

Answer: C.
Show more

Hi Bunuel,

Because when a 3 member committee is formed from a 6 member committee automatically two3 member committee are formed. Also, doing this way we would double count the 3 member committees in such a scenario. So should the answer be \(\frac{6C3*3C3}{2!}=10\).

Could you please help me with this fundamental query.
Thanks in advance! :-)
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If a 3-member subcommittee is to be formed from a certain 6- 27 Jul 2020, 15:25
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I'm not Bunuel but I can try to explain.

Out of a 6-member committee, we are selecting 3 members for a subcommittee. Therefore for the 1st selection for this subcommittee we have a total of 6 different members to choose from. Then for the 2nd selection, we have 5 members to choose from, and then the 3rd selection we have 4 members to choose from. We don't distinguish the order that we make these selections since the members are not uniquely identifiable, so we have to make sure to remove any duplicate ways to select these 3 members. There is a couple ways you can conceptualize this:

Option 1: Use Combinations formula with [total # of selections]!/[total # of selection - # we are selecting]! * [# we are selecting]!
\(C\binom 3 6\) = \(\frac{6!}{(6−3)!3!}\)= 20

Option 2: Count out the total number of selections per each member and divide by the factorial of the # of members you are selecting
\(\frac{(6)(5)(4)}{3!}\)

SidJainGMAT
Bunuel
goodyear2013
if a 3-member subcommittee is to be formed from a certain 6-member committee, how many different such subcommittee are possible?

A) 6
B) 18
C) 20
D) 108
E) 216

Choosing 3 out of 6: \(C^3_6=\frac{6!}{(6-3)!3!}=20\).

Answer: C.

Hi Bunuel,

Because when a 3 member committee is formed from a 6 member committee automatically two3 member committee are formed. Also, doing this way we would double count the 3 member committees in such a scenario. So should the answer be \(\frac{6C3*3C3}{2!}=10\).

Could you please help me with this fundamental query.
Thanks in advance! :-)
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If a 3-member subcommittee is to be formed from a certain 6- 17 Aug 2021, 21:39
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A question on combination is a question on selection.

So 3 people can be selected from 6 in 6C3 ways = 6!/3!*3!

= 20 ways.

(option c)

D.S
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If a 3-member subcommittee is to be formed from a certain 6- 31 Dec 2023, 04:41
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Bunuel
goodyear2013
if a 3-member subcommittee is to be formed from a certain 6-member committee, how many different such subcommittee are possible?

A) 6
B) 18
C) 20
D) 108
E) 216

Choosing 3 out of 6: \(C^3_6=\frac{6!}{(6-3)!3!}=20\).

Answer: C.
Show more

Hi Bunuel, why is combination method of 6C3 works here even though question state how many "different" such subcommittee are possible? Is this just coindidence with 6x5x4/3! = 20.
Wouldn't the combination method including the duplicates here?
Could you kindly help clarify? Thanks
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If a 3-member subcommittee is to be formed from a certain 6- 31 Dec 2023, 06:15
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Kimberly77
Bunuel
goodyear2013
if a 3-member subcommittee is to be formed from a certain 6-member committee, how many different such subcommittee are possible?

A) 6
B) 18
C) 20
D) 108
E) 216

Choosing 3 out of 6: \(C^3_6=\frac{6!}{(6-3)!3!}=20\).

Answer: C.

Hi Bunuel, why is combination method of 6C3 works here even though question state how many "different" such subcommittee are possible? Is this just coindidence with 6x5x4/3! = 20.
Wouldn't the combination method including the duplicates here?
Could you kindly help clarify? Thanks
Show more

The formula \(C^n_k=\frac{n!}{k!(n-k)!}\) gives the number of different groups of k elements that can be selected from a set of n distinct elements. For instance, consider the set {a, b, c}. The formula \(C^2_3=\frac{3!}{2!(3-2)!}=3\) gives the number of 2-element groups that can be formed from these 3 elements: {a, b}, {a, c}, and {b, c}.

21. Combinatorics/Counting Methods



For more:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread


Hope it helps.
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If a 3-member subcommittee is to be formed from a certain 6- 02 Jan 2024, 13:11
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Thanks Bunuel, I will see if I am able to figure it out.
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