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

Question

Fool-proof method to Differentiate between Permutation & Combination Questions - Exercise Question #1

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.

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

Learn to use the Keyword Approach in Solving PnC question from the following article:

Article-1:      Learn when to “Add” and “Multiply” in Permutation & Combination questions

Article-2:      Fool-proof method to Differentiate between Permutation & Combination Questions

EgmatQuantExpert · Answer

Solution

Given:

• We need to select a cricket team i.e. 11 players
 o We need to select 5 batsmen from 7 batsmen and,
o We need to select 4 bowlers from 6 bowlers and,
o We need to select 1 wicket keeper from 3 wicket keepers and,
o We need to select 1 all-rounder from 3 all-rounders.

To find:  
• The number of ways we can from a cricket team out of 7 batsmen, 6 bowlers, 3 allrounders and 3 wicketkeepers available.

Approach and Working:

Since we need to select 5 batsmen, 4 bowlers, and 1 wicket keeper and 1 all-rounder in the cricket team,
 • Hence, number of ways to form the cricket team= ways to select 5 batsmen from 7 batsmen* ways to select 4 bowlers from 6 bowlers * ways to select 1 wicket keeper from 3 wicket keepers * ways to select 1 all-rounder from 3 all-rounders

• Thus, total ways=  ^7c_5 * ^6c_4 * ^3c_1 * ^3c_1

• =21*15*3*3= 2835

Hence, in 2835 ways we can select a cricket team.

Hence, option D is the correct answer.

Answer: D

Wildflower · Answer

7C5 * 6C4 * 3C1 * 3C1 = 7!/(2!5!) * 6!/(2!4!) * 3!/(2!1!) * 3!/(2!1!) = 21*15*3*3 = 2835 --> D

imo, answer is option D.

balaji4799 · Answer

A simple PS question where one has to choose 5 batsman from 7 batsman, 4 bowlers out of 6 bowlers, 1 wicket keeper out of 3 wicket keepers and 1 all rounder from 3 allrounders. So according to combinations principles, the number of ways we can choose r items from n items is found out as nCr which equals to n!/((n-r)!*r!). So here the solution is , 7C5*6C4*3C1*3C1 = 2835 (D)

EgmatQuantExpert · Answer

Hey everyone, The official solution to the question has been posted.

Regards, Ashutosh e-GMAT

push12345 · Answer

It's a simple question of selection for which we use combination formula

7C5*6C4*3C1*3C1

Posted from my mobile device

BrentGMATPrepNow · Answer

Take the task of creating a cricket team and break it into   stages   and then apply the  Fundamental Counting Principle

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. So, be sure to learn it.

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bumpbot · Answer

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