It is currently 23 Sep 2017, 09:58

Happening Now:

Alleviate MBA app anxiety! Come to Chat Room #2

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

Hide Tags

Manager
Joined: 03 Feb 2010
Posts: 68

Kudos [?]: 147 [0], given: 4

Consider the number of way a committee of 3 can be selected [#permalink]

Show Tags

20 Apr 2010, 12:09
1
This post was
BOOKMARKED
00:00

Difficulty:

(N/A)

Question Stats:

100% (01:50) correct 0% (00:00) wrong based on 8 sessions

HideShow timer Statistics

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.

Kudos [?]: 147 [0], given: 4

Math Expert
Joined: 02 Sep 2009
Posts: 41698

Kudos [?]: 124616 [1], given: 12079

Show Tags

20 Apr 2010, 12:28
1
KUDOS
Expert's post
ksharma12 wrote:
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.

What is the justification?

$$C^3_7-C^1_5=30$$

$$C^3_7$$ - # 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 $$C^1_5$$).
_________________

Kudos [?]: 124616 [1], given: 12079

Manager
Joined: 03 Feb 2010
Posts: 68

Kudos [?]: 147 [0], given: 4

Show Tags

20 Apr 2010, 13:27
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?

Kudos [?]: 147 [0], given: 4

Math Expert
Joined: 02 Sep 2009
Posts: 41698

Kudos [?]: 124616 [1], given: 12079

Show Tags

20 Apr 2010, 13:42
1
KUDOS
Expert's post
ksharma12 wrote:
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?

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 $$C^1_5=5$$.

ksharma12 wrote:
Essentially the same thing as at least 1 problem?

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

Kudos [?]: 124616 [1], given: 12079

Senior Manager
Joined: 27 May 2012
Posts: 390

Kudos [?]: 85 [0], given: 477

Show Tags

10 Sep 2012, 00:11
Bunuel wrote:
ksharma12 wrote:
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.

What is the justification?

$$C^3_7-C^1_5=30$$

$$C^3_7$$ - # 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 $$C^1_5$$).

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?
_________________

- Stne

Kudos [?]: 85 [0], given: 477

Director
Joined: 22 Mar 2011
Posts: 611

Kudos [?]: 1044 [1], given: 43

WE: Science (Education)

Show Tags

10 Sep 2012, 04:11
1
KUDOS
stne wrote:
Bunuel wrote:
ksharma12 wrote:
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.

What is the justification?

$$C^3_7-C^1_5=30$$

$$C^3_7$$ - # 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 $$C^1_5$$).

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?

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!...
_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

Kudos [?]: 1044 [1], given: 43

Senior Manager
Joined: 27 May 2012
Posts: 390

Kudos [?]: 85 [0], given: 477

Show Tags

10 Sep 2012, 07:27
Thanks for the reassurance
_________________

- Stne

Kudos [?]: 85 [0], given: 477

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 17620

Kudos [?]: 271 [0], given: 0

Re: Consider the number of way a committee of 3 can be selected [#permalink]

Show Tags

22 Jan 2014, 16:31
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 271 [0], given: 0

Moderator
Joined: 20 Dec 2013
Posts: 185

Kudos [?]: 74 [0], given: 71

Location: United States (NY)
GMAT 1: 640 Q44 V34
GMAT 2: 710 Q48 V40
GMAT 3: 720 Q49 V40
GPA: 3.16
WE: Consulting (Venture Capital)
Re: Consider the number of way a committee of 3 can be selected [#permalink]

Show Tags

22 Jan 2014, 17:26
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!
_________________

Kudos [?]: 74 [0], given: 71

Math Expert
Joined: 02 Sep 2009
Posts: 41698

Kudos [?]: 124616 [0], given: 12079

Re: Consider the number of way a committee of 3 can be selected [#permalink]

Show Tags

23 Jan 2014, 04:25
m3equals333 wrote:
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!

_________________

Kudos [?]: 124616 [0], given: 12079

Re: Consider the number of way a committee of 3 can be selected   [#permalink] 23 Jan 2014, 04:25
Similar topics Replies Last post
Similar
Topics:
65 If a committee of 3 people is to be selected from among 5 31 25 Jan 2017, 10:18
156 If a committee of 3 people is to be selected from among 5 35 25 Jul 2017, 07:21
35 If a committee of 3 people is to be selected from among 5 16 07 Jul 2013, 14:40
4 In how many ways can a group of 3 people be selected 5 01 Sep 2016, 03:32
23 In how many ways can 3-digit numbers be formed selecting 3 d 23 19 Aug 2017, 06:12
Display posts from previous: Sort by