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In how many ways a cricket team consisting of 5 batsmen, 4

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by AAPL » Fri Feb 22, 2019 4:02 am
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e-GMAT

In how many ways a cricket team consisting of 5 batsmen, 4 bowlers, and 1 wicketkeeper and 1 all-rounder can be selected from the pool of 7 batsmen, 6 bowlers, 3 wicketkeepers and 3 allrounders?

A. 567
B. 1420
C. 2256
D. 2835
E. 5670

OA D
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Source: — Problem Solving |

by GMATGuruNY » Fri Feb 22, 2019 5:12 am
AAPL wrote:e-GMAT

In how many ways a cricket team consisting of 5 batsmen, 4 bowlers, and 1 wicketkeeper and 1 all-rounder can be selected from the pool of 7 batsmen, 6 bowlers, 3 wicketkeepers and 3 allrounders?

A. 567
B. 1420
C. 2256
D. 2835
E. 5670
From 7 batsmen, the number of ways to choose 5 = 7C5 = (7*6*5*4*3)/(5*4*3*2*1) = 21.
From 6 bowlers, the number of ways to choose 4 = 6C4 = (6*5*4*3)/(4*3*2*1) = 15.
From 3 wicketkeepers, the number of ways to choose 1 = 3.
From 3 allrounders, the number of ways to choose 1 = 3.
To combine these options, we multiply:
21*15*3*3 = integer with a units digit of 5.

The correct answer is D.
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by Brent@GMATPrepNow » Fri Feb 22, 2019 6:06 am
AAPL wrote:e-GMAT

In how many ways a cricket team consisting of 5 batsmen, 4 bowlers, and 1 wicketkeeper and 1 all-rounder can be selected from the pool of 7 batsmen, 6 bowlers, 3 wicketkeepers and 3 allrounders?

A. 567
B. 1420
C. 2256
D. 2835
E. 5670

OA D
Take the task of creating a cricket team and break it into stages.

Stage 1: Select 5 batsmen
Since the order in which we select the batsmen does not matter, we can use combinations.
We can select 5 batsmen from 7 batsmen in 7C5 ways (21 ways)
So, we can complete stage 1 in 21 ways

Stage 2: Select 4 bowlers
Since the order in which we select the bowlers does not matter, we can use combinations.
We can select 4 bowlers from 6 bowlers in 6C2 ways (15 ways)
So, we can complete stage 2 in 15 ways

Stage 3: Select 1 wicketkeeper
There are 3 wicketkeepers from which to choose, so we can complete this stage in 3 ways.

Stage 4: Select 1 all-rounder
There are 3 all-rounders from which to choose, so we can complete this stage in 3 ways.

By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus create a cricket team) in (21)(15)(3)(3) ways ( = 2835 ways)

Answer: D
--------------------------

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat- ... /video/775

You can also watch a demonstration of the FCP in action: https://www.gmatprepnow.com/module/gmat ... /video/776

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should ... 67256.html
- https://www.beatthegmat.com/counting-pr ... 44302.html
- https://www.beatthegmat.com/picking-a-5 ... 73110.html
- https://www.beatthegmat.com/permutation ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html


MEDIUM
- https://www.beatthegmat.com/combinatori ... 73194.html
- https://www.beatthegmat.com/arabian-hor ... 50703.html
- https://www.beatthegmat.com/sub-sets-pr ... 73337.html
- https://www.beatthegmat.com/combinatori ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-se ... 71047.html
- https://www.beatthegmat.com/combinatori ... 67079.html


DIFFICULT
- https://www.beatthegmat.com/wonderful-p ... 71001.html
- https://www.beatthegmat.com/permutation ... 73915.html
- https://www.beatthegmat.com/permutation-t122873.html
- https://www.beatthegmat.com/no-two-ladi ... 75661.html
- https://www.beatthegmat.com/combinations-t123249.html


Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Scott@TargetTestPrep » Sun Feb 24, 2019 5:31 am
AAPL wrote:e-GMAT

In how many ways a cricket team consisting of 5 batsmen, 4 bowlers, and 1 wicketkeeper and 1 all-rounder can be selected from the pool of 7 batsmen, 6 bowlers, 3 wicketkeepers and 3 allrounders?

A. 567
B. 1420
C. 2256
D. 2835
E. 5670

OA D
The 5 batsmen can be chosen in 7C5 ways. The 4 bowlers can be chosen in 6C4 ways. One wicketkeeper and one all-rounder each can be chosen in 3C1 ways. In total, the team can be chosen in

7C5 x 6C4 x 3C1 x 3C1 = 7!/(5!2!) x 6!/(4!2!) x 3 x 3 = [(7 x 6)/2] x [(6x5)/2] x 3 x 3

= 7 x 3 x 3 x 5 x 3 x 3 = 2835

ways.

Answer: D

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