If a 3-member subcommittee is to be formed from a certain 6-

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

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

Answer: C.
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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!

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
\(\frac{(6)(5)(4)}{3!}\)

\(6C_3 = 20\)

Thus, the answer must be (C) 20
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The number of possible subcommittees is 6C3 = (6 x 5 x 4)/3! = 20.

Answer: C
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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
GMAT SME
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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

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}.

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Hope it helps.
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Thanks Bunuel, I will see if I am able to figure it out.