It is currently 24 Mar 2018, 18:24

### 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

# Probability Club

Author Message
Manager
Joined: 29 May 2008
Posts: 110

### Show Tags

22 Jun 2009, 11:24
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

The X fraternity must choose a delegation of three senior members and two junior members for an annual intrafaternity conference. If X fraternity has 12 Senior members and 11 Junior members, How many different delegations are possible?

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
Current Student
Joined: 03 Aug 2006
Posts: 112

### Show Tags

22 Jun 2009, 12:05
This is a combinations problem.

$$C^n_k = \frac{n!}{k!(n-k)!}$$

where $$n$$ = total number of items available to select and $$k$$ = number of items to be selected.

In this problem there are two distinct sets: 1 set of senior members and 1 set of junior members.

The number of ways we can select 3 ($$k$$) senior members from a total of 12 ($$n$$) senior members:

$$\Rightarrow C^n_k = \frac{n!}{k!(n-k)!}$$

$$\Rightarrow C^{12}_3 = \frac{12!}{3!(12-3)!}$$

$$\Rightarrow C^{12}_3 = \frac{12!}{3!9!}$$

$$\Rightarrow C^{12}_3 = \frac{12 \times 11 \times 10}{3 \times 2} = 220$$

Similarly the number of ways we can select 2 ($$k$$) junior members from a total of 11 ($$n$$) junior members:

$$\Rightarrow C^n_k = \frac{n!}{k!(n-k)!}$$

$$\Rightarrow C^{11}_2 = \frac{11!}{2!(11-2)!}$$

$$\Rightarrow C^{11}_2 = \frac{11!}{2!9!}$$

$$\Rightarrow C^{11}_2 = \frac{11 \times 10}{2} = 55$$

Total number of different delegations of senior and junior members possible:

$$= 220 \times 55$$

$$= 12100$$
Manager
Joined: 15 Apr 2008
Posts: 159

### Show Tags

22 Jun 2009, 12:06
12C3+11C2

220+55=275

Intrafraternity consists of 3 senior members from a pool of 12 senior members, so 12C3 and 2 junior members from a pool of 11 members so 11C2.

we add both the combinations to get 275 different types of delegations.
Intern
Joined: 04 Mar 2008
Posts: 42

### Show Tags

22 Jun 2009, 14:58
The answer is 220*55 = 12100.

what is OA?
Manager
Joined: 29 May 2008
Posts: 110

### Show Tags

23 Jun 2009, 07:45
nookway

You are right

thanks a lot
Senior Manager
Joined: 23 Jun 2009
Posts: 355
Location: Turkey
Schools: UPenn, UMich, HKS, UCB, Chicago

### Show Tags

23 Jun 2009, 07:50
The answer is 220*55 = 12100.

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
Re: Probability Club   [#permalink] 23 Jun 2009, 07:50
Display posts from previous: Sort by

# Probability Club

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.