A group of medical interns at Bohemus Medical School want to go on dat

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:

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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The first girl can go on a date in 5 ways.
The second can go on a date in 4 ways.
The third in 3 and so on.

This is equal to 5! i.e. 120.

Hence, the answer should be (E).

I would appreciate a kudos if you liked my solution!

Because they are all straight, there are five pairs possible.

Total arrangements of these pairs:
5! = 120

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\(?\,\,\,\,:\,\,\,\,\# \,\,\,{\text{guy - girl}}\,\,{\text{pairs}}\)

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.

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