Lucas C wrote:A certain city contains 25 consulates, each of which employs 25 people. If an international research task force of 24 people is to be chosen from these people, none of whom can come from the same consulate, what is the greatest number of distinct task force teams that could be formed?
a) 24^25
b) 25^24
c) 25^25
d) 24(25^24)
e) 25(24^25)
Answer: C
Take the task of creating the international research task force and break it into
stages.
Stage 1: Select the 24 consulates (from which we will select 1 employee from each)
Since the order in which we select the consulates does not matter, we can use combinations.
We can select 24 consulates from the 25 consulates women in 25C24 ways (25 ways)
So, we can complete stage 1 in
25 ways
NOTE: this is the same as selecting 1 consulate to NOT have an employee chosen. That is, 25C1 = 25C24 = 25
Stage 2: From 1 of the chosen consulates, select 1 employee
Since there are 25 employees in a consulate, we can complete this stage in
25 ways.
Stage 3: From 1 of the REMAINING consulates chosen, select 1 employee
Since there are 25 employees in a consulate, we can complete this stage in
25 ways.
Stage 4: From 1 of the REMAINING consulates chosen, select 1 employee
Since there are 25 employees in a consulate, we can complete this stage in
25 ways.
Stage 5: From 1 of the REMAINING consulates chosen, select 1 employee
Since there are 25 employees in a consulate, we can complete this stage in
25 ways.
.
.
.
.
Since we choose 24 consulates in stage 1, we will have to make 24 employee selections (1 from each chosen consulate).
.
.
.
Stage 24: From 1 of the REMAINING consulates chosen, select 1 employee
Since there are 25 employees in a consulate, we can complete this stage in
25 ways.
Stage 25: From the last REMAINING consulate chosen, select 1 employee
Since there are 25 employees in a consulate, we can complete this stage in
25 ways.
By the Fundamental Counting Principle (FCP), we can complete all 25 stages (and thus create the create the international research task force) in
(25)(25)(25)(25)(25)(25)(25)(25)(25)(25)(25)(25)(25)(25)(25)(25)(25)(25)(25)(25)(25)(25)(25)(25)(25) ways ([spoiler]= 25^25 ways[/spoiler])
Answer:
C
--------------------------
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
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https://www.beatthegmat.com/counting-pro ... 44302.html
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https://www.beatthegmat.com/picking-a-5- ... 73110.html
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https://www.beatthegmat.com/permutation- ... 57412.html
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https://www.beatthegmat.com/simple-one-t270061.html
MEDIUM
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https://www.beatthegmat.com/combinatoric ... 73194.html
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https://www.beatthegmat.com/arabian-hors ... 50703.html
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https://www.beatthegmat.com/sub-sets-pro ... 73337.html
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https://www.beatthegmat.com/combinatoric ... 73180.html
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https://www.beatthegmat.com/digits-numbers-t270127.html
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https://www.beatthegmat.com/doubt-on-sep ... 71047.html
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https://www.beatthegmat.com/combinatoric ... 67079.html
DIFFICULT
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https://www.beatthegmat.com/wonderful-p- ... 71001.html
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https://www.beatthegmat.com/permutation- ... 73915.html
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https://www.beatthegmat.com/permutation-t122873.html
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https://www.beatthegmat.com/no-two-ladie ... 75661.html
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https://www.beatthegmat.com/combinations-t123249.html
Cheers,
Brent