Consider the number of way a committee of 3 can be selected

Question

Consider the number of way a committee of 3 can be selected from 7 people- A, B, C, D, E, F, G if order does not matter, and C and E cannot be chosen together.

Bunuel · Answer

C^3_7-C^1_5=30

C^3_7  - # of ways we can choose any 3 out of 7 (without restriction).
 C^1_5  - # of groups with C and E together (if C and E are in chosen group, then third member can be any out of 5 left, so total # of groups is  C^1_5 ).

ksharma12 · Answer

When you say C(1,5) do you mean 1x1xC(1,5)?

Because only 1 way to choose C and 1 Way to choose E?

Therefore you minus that from the complete total ways to pick 3 from 7?

Essentially the same thing as at least 1 problem?

Bunuel · Answer

Yes. You can write this as  C^1_1*C^1_1*C^1_5=5  (one way to choose C and one way to choose choose E) OR  C^2_2*C^1^5=5  (one way to choose C and E, from C and E), which is basically  C^1_5=5 .

If you mean in a way: total-opposite, then yes.

stne · Answer

just for discussion sake and to clear the concept I was wondering what would happen if the order mattered in the above scenario.
Eva ..Bunuel and others in the community request you to please contribute .

lets assume ABCDEFG is a 7 letter word so ABC is different from CAB meaning order matters 
so if I wanted to form 3 letter words such that CE should not be together then , what would be the number of ways this could be done ?

I tried to a certain extent please verify if this is correct or not.

C^7_3 * 3! - C^1_1 *C^1_1*C^5_1*3! =180

C^7_3  = # of ways to select 3 out of 7 
3! = # of ways to arrange them among themselves as order matters

C^1_1 = choosing c
 C^1_1  = choosing e 
 C^5_1 = # of ways to select one from remaining 5 , after C and E have been chosen in the community.
 3!  = # of to arrange the 3 letter committee containing both C and E

Hope this is correct?

EvaJager · Answer

Yes, this is correct. I just have my own preferences to count...so, for example, choose 3 out of 7 when order matters I write directly 7*6*5, meaning I directly take into account the order (why write the formula with the factorials for nCk and then multiply by k!, reduce...). First choice 7 options, second choice 6, third 5. And I am not even writing the 1C1 factors. There is nothing wrong with it, but I know that C and E must be chosen, then I need just one extra person (letter

), so I can choose 1 out of 5, for which again I am not writing the 5C1, and then having C, E and * (somebody), I have to consider all the permutations of the three, so 3!...

stne · Answer

Thanks for the reassurance

m3equals333 · Answer

Bunuel, can we also think of it as 6C3 + 6C3 - 3C5?

I am trying to calculate combinations each C and E with other 5 separately, adding them together, then subtracting out one of the double counted 3 team combinations of the other 5 members. Not sure if this is a proper approach? Thanks!

Bunuel · Answer

Your approach is correct.

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

Consider the number of way a committee of 3 can be selected

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

Consider the number of way a committee of 3 can be selected

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards