A company consists of 5 senior and 3 junior staff officers. If a

Question

A company consists of 5 senior and 3 junior staff officers. If a committee is created with 3 senior and 1 junior staff officers, in how many ways can the committee be formed?

(A) 12
(B) 30
(C) 45
(D) 80
(E) 200

Kudos for a correct  solution .

camlan1990 · Answer

Choose 3 senior from 5 senior and choose 1 junior from 3 junior:

3C5 * 1C3 = 10*3 = 30

Ans: B

BrentGMATPrepNow · Answer

Take the task of creating the committee and break it into stages . Stage 1 : Select 3 senior officers for the committee Since the order in which we select the senior officers does not matter, we can use combinations. We can select 3 senior officers from 5 senior officers in 5C3 ways (10 ways) So, we can complete stage 1 in 10 ways If anyone is interested, we have a free video on calculating combinations (like 5C3) in your head: https://www.gmatprepnow.com/module/gmat- ... /video/789 Stage 2 : Select 1 junior officer for the committee There are 3 junior officers to choose from, so we can complete this stage in 3 ways. By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus create a 4-person committee) in (10)(3) ways (= 30 ways) Answer: B 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-counting/video/775

ShristiK · Answer

Another way to approach this problem is as follows:

We need to choose 3 seniors from a set of 5 seniors: _ * _ * _
Each of the seniors can be chosen in x=  5  *  4  *  3  ways.
Here, the order in which we choose the seniors doesn't matter; so, we divide x by 3! (for the 3 seats).
So we get ( 5*4*3 )/ 3! = 10

Next, we need to choose 1 junior from a set of 3 juniors: _
Each junior can be chosen in  3  ways

Since these are two separate events, we will multiply them to get our answer.
The members can be chosen in 10 * 3 = 30 ways.

icefrog · Answer

Answer is 30.

Say there are three positions _ _ _ for the senior officers and 1 position _ for the junior officer.

The first _ can be filled by 5, the second _ by 4 ( since the same person cannot occupy the second one as well) and the third _ by 3.
So 5*4*3 are possible. But since, we are not interested in the order, we have to divide it by 3!. = 10.
The junior officer position can be filled in 3 ways. So total number of ways the committee can be formed is 3*10= 30

Hemavakade · Answer

We are choosing a 4 member committee. We are not concerned about the order in which they are chosen..

3 seniors can be chosen from 5 seniors in 5C3 ways = 10

1 junior can be chosen from 3 juniors in 3C1 ways = 3

Using the Fundamental counting principle we can choose the committee in 10 x 3 ways = 30

Answer = B

marinnesquick · Answer

5C3 * 3C1 = (5! / 2!3!) * (3!/2!) = 30 (B)

happyface101 · Answer

A company consists of 5 senior and 3 junior staff officers. If a committee is created with 3 senior and 1 junior staff officers, in how many ways can the committee be formed?

Step 1: recognize how many different groups are there on the committee? there are 2, a senior group and a junior group

Step 2: determine how many ways to select within each group. Senior: 3 senior are on the committee, there are a total of 5 seniors, so need to choose 3 among 5 seniors. That's 5 choose 3 = 5!/(3!2!) = 10. Junior: 1 junior on the committee, there are a total of 3 juniors, so need to choose 1 among 3 juniors. That's 3 choose 1 = 3!/(1!2!) = 3

Step 3: multiply the ways to select from each group. Senior: 10 ways. Junior: 3 ways. 10*3= 30 ways. >> answer B.

Please give me a Kudo if this is helpful - thanks so much!

longfellow · Answer

Answer is 30 calculated by 5c3*3c1=30

dabaobao · Answer

Official Solution  
Credit: Veritas Prep

You have to select 3 senior and 1 junior officers. Note here that you don’t have to arrange them in any way. You just have to select.

There are a total of 5 senior officers. You can select 3 of them in (5∗4∗3)/3! ways. Note that we divide by 3! to un-arrange.

There are 3 junior officers and you have to select one of them. You can do that in 3 different ways. Note here that you don’t need to do any calculations when you have to select just one person. Out of 3 people (say A, B and C), you can select one in 3 ways (you can select A or B or C).

So you can select 3 senior and 1 junior officers in (5∗4∗3/3!)∗3=30 ways

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