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A club has 10 members. One president and two vice-presidents are elect

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A club has 10 members. One president and two vice-presidents are elect  [#permalink]

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New post 28 Nov 2018, 01:09
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[Math Revolution GMAT math practice question]

A club has 10 members. One president and two vice-presidents are elected. In how many ways can they be selected?

A. 240
B. 280
C. 300
D. 320
E. 360

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A club has 10 members. One president and two vice-presidents are elect  [#permalink]

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New post 28 Nov 2018, 02:10
2
MathRevolution wrote:
[Math Revolution GMAT math practice question]

A club has 10 members. One president and two vice-presidents are elected. In how many ways can they be selected?

A. 240
B. 280
C. 300
D. 320
E. 360



There are two ways of doing.

Either you select one president and then select 2 Vice presidents from remaining(that is 10-1=9 remaining) --10*
or
You select 2 vice president and then select 1 president from remaining(that is 10-2= 9 remaining) --10C2*8C1

For first way --

No of ways of selecting one President from 10 = 10C1 = 10 ways
No of ways of selecting two Vice- President from remaining (9) = 9C2
Total number of ways = 10*9C2= 360
So E is the answer.


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Re: A club has 10 members. One president and two vice-presidents are elect  [#permalink]

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New post 28 Nov 2018, 06:43
Top Contributor
MathRevolution wrote:
[Math Revolution GMAT math practice question]

A club has 10 members. One president and two vice-presidents are elected. In how many ways can they be selected?

A. 240
B. 280
C. 300
D. 320
E. 360


Take the task of selecting the president and two vice-presidents and break it into stages.

Stage 1: Select the president
There are 10 people to choose from. So, we can complete stage 1 in 10 ways

Stage 2: Select two people to be the vice-presidents
Since the order in which we select the two people does not matter, we can use combinations.
We can select 2 people from the remaining 9 people in 9C2 ways (36 ways)
So, we can complete stage 2 in 36 ways

If anyone is interested, we have a free video on calculating combinations (like 9C2) in your head: http://www.gmatprepnow.com/module/gmat- ... /video/789

By the Fundamental Counting Principle (FCP), we can complete the two stages (and thus select a president and two vice presidents) in (10)(36) ways (= 360 ways)

Answer: E

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

RELATED VIDEO FROM OUR COURSE

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CEO
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Joined: 11 Sep 2015
Posts: 3237
Location: Canada
Re: A club has 10 members. One president and two vice-presidents are elect  [#permalink]

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New post 28 Nov 2018, 06:46
Top Contributor
MathRevolution wrote:
[Math Revolution GMAT math practice question]

A club has 10 members. One president and two vice-presidents are elected. In how many ways can they be selected?

A. 240
B. 280
C. 300
D. 320
E. 360


Take the task of selecting the president and two vice-presidents and break it into stages.

Stage 1: Select the president
There are 10 people to choose from. So, we can complete stage 1 in 10 ways

Stage 2: Select two people to be the vice-presidents
Since the order in which we select the two people does not matter, we can use combinations.
We can select 2 people from the remaining 9 people in 9C2 ways (36 ways)
So, we can complete stage 2 in 36 ways

If anyone is interested, we have a free video on calculating combinations (like 9C2) in your head: http://www.gmatprepnow.com/module/gmat- ... /video/789

By the Fundamental Counting Principle (FCP), we can complete the two stages (and thus select a president and two vice presidents) in (10)(36) ways (= 360 ways)

Answer: E

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

RELATED VIDEO FROM OUR COURSE

_________________

Test confidently with gmatprepnow.com
Image

Math Revolution GMAT Instructor
User avatar
V
Joined: 16 Aug 2015
Posts: 6629
GMAT 1: 760 Q51 V42
GPA: 3.82
Premium Member
Re: A club has 10 members. One president and two vice-presidents are elect  [#permalink]

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New post 30 Nov 2018, 01:18
=>

The number of ways to choose one president out of \(10\) people is 10C1 = 10.
The number of ways to choose two vice-presidents out of the remaining \(9\) people is 9C2 = \(\frac{(9*8)}{(1*2)} = 36.\)
Thus, the number of ways to choose one president and two vice-presidents out of \(10\) people is \(10 * 36 = 360.\)

Therefore, the answer is E.
Answer: E
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Re: A club has 10 members. One president and two vice-presidents are elect &nbs [#permalink] 30 Nov 2018, 01:18
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