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A group of medical interns at Bohemus Medical School want to

by subh2273 » Mon Oct 08, 2018 9:41 pm

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A group of medical interns at Bohemus Medical School want to go on dates. There are 5
girls and 5 guys. Assuming girls go on dates with guys, how many possible ways can these
10 medical interns date each other?

(A)
10

(B)
25

(C)
60

(D)
90

(E)
120
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by GMATGuruNY » Tue Oct 09, 2018 10:10 am
subh2273 wrote:A group of medical interns at Bohemus Medical School want to go on dates. There are 5
girls and 5 guys. Assuming girls go on dates with guys, how many possible ways can these
10 medical interns date each other?

(A)
10

(B)
25

(C)
60

(D)
90

(E)
120
In other words:
How many ways can 5 girls and 5 guys be divided into pairs, if each pair must consist of one girl and one guy?

Number of options for the first girl = 5. (Any of the 5 guys.)
Number of options for the second girl = 4. (Any of the 4 remaining guys).
Number of options for the third girl = 3. (Any of the 3 remaining guys.)
Number of options for the fourth girl = 2. (Either of the 2 remaining guys.)
Number of options for the fifth girl = 1. (Only one guy left.)
To combine the options above, we multiply:
5*4*3*2*1 = 120.

The correct answer is E.
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A group of medical interns at Bohemus Medical School want to

by Brent@GMATPrepNow » Tue Oct 09, 2018 10:20 am
subh2273 wrote:A group of medical interns at Bohemus Medical School want to go on dates. There are 5
girls and 5 guys. Assuming girls go on dates with guys, how many possible ways can these
10 medical interns date each other?

(A)
10

(B)
25

(C)
60

(D)
90

(E)
120
Take the task of arranging dates and break it into stages.

Let A, B, C, D and E represent the 5 girls

Stage 1: Select a boy to date girl A
We can choose any of the 5 boys, so we can complete stage 1 in 5 ways

Stage 2: Select a boy to date girl B
Since we already selected a boy in stage 1, there are 4 boys remaining to choose from.
So we can complete stage 2 in 4 ways

Stage 3: Select a boy to date girl C
There are 3 boys remaining to choose from.
So we can complete stage 3 in 3 ways

Stage 4: Select a boy to date girl D
2 boys remaining. So we can complete stage 4 in 2 ways

Stage 5: Select a boy to date girl E
There is 1 boy remaining to be seated, so we can complete stage 5 in 1 way

By the Fundamental Counting Principle (FCP), we can complete all 5 stages (and thus arrange all dates) in (5)(4)(3)(2)(1) ways (= 120 ways)

Answer: E

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by fskilnik@GMATH » Tue Oct 09, 2018 10:46 am
subh2273 wrote:A group of medical interns at Bohemus Medical School want to go on dates. There are 5
girls and 5 guys. Assuming girls go on dates with guys, how many possible ways can these
10 medical interns date each other?
$$(A) 10 \,\,\,\,\,\,\,\,\,\,\, (B) 25 \,\,\,\,\,\,\,\,\,\,\, (C) 60 \,\,\,\,\,\,\,\,\,\,\, (D) 90 \,\,\,\,\,\,\,\,\,\,\, (E) 120$$
\[?\,\,\,\,:\,\,\,\,\# \,\,\,{\text{guy - girl}}\,\,{\text{pairs}}\]
Imagine girls in a row (say) in alphabetical order.

FOCUS: number of ways to put 5 guys in a row (parallel to the first row, each guy facing one girl - this is a pair!)
\[?\, = \,{P_{\,5}}\,\, = 5! = 120\]

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
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by Scott@TargetTestPrep » Fri Oct 12, 2018 6:57 am
subh2273 wrote:A group of medical interns at Bohemus Medical School want to go on dates. There are 5
girls and 5 guys. Assuming girls go on dates with guys, how many possible ways can these
10 medical interns date each other?

(A)
10

(B)
25

(C)
60

(D)
90

(E)
120
The first girl has 5 choices of guys, the second girl has 4 choices of guys (after a guy is picked by the first girl), the third girl has 3 choices of guys (after a guy is picked by the second girl), and so on. So the number of ways these 10 interns can date each other is 5 x 4 x 3 x 2 x 1 = 120.

Answer: E

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