A certain travel agency organizes a 5-city sightseeing package from a

Hey Bunuel, could you please explain why is it 252?
Are we ignoring the selection by 2 people... and taking it only as one's choice i.e. 10C5? as the OA answer says 252 (10C5).

KarishmaB

Can you please help in this question?

I thought that because 2 different people are selecting the answer should be 10C3 * 7C2. But instead answer is 10C5. Why is that?

We want to know how many groups of 5 cities could be chosen. It doesn't matter who chooses them -- there are 10C5 = 252 groups of five cities you can choose from ten cities.

10C3 * 7C2 is the answer to a different question -- that's the answer to the question "in how many ways can the client pick 3 cities and the travel agent then pick 2?" or something similar, where we distinguish the cities selected by the client and those selected by the agent, but the question doesn't ask us to do that.

There are 10 cities and you have to make a package of 5. In how many ways can you choose 5 cities out of 10? In 10C5.
So say cities are A, B, C, D, E, F, G, H, I, J

One package would be A, B, C, D, H
Another would be C, D, F, H, A
etc...
Is there any distinction between the cities? No. We are told we don't need to worry about the arrangement.
Say you and me have to select 5 cities together. You select 3 and I pick another 2. We have 5 selections. Does it change if you pick 1 and I pick 4? No. The selection is still the same - same number picked from the same list. We get the same selections.

Think what 10C3 * 7C2 is: If I have a group of 10 cities and I pick 3 out of them, I get 10C3. If I have another 7 cities and I pick 2 of them I get 7C2.
Then I get 10C3*7C2 as my selection of 5 cities out of 17 cities from two different groups - Group 1 of 10 cities and group 2 of 7 cities such that I pick 3 from group 1 and 2 from group 2.

Or if I want to visit 3 cities this year and 2 cities next year (without repeating) from the list of 10 cities, I can select 1 group of 3 cities in 10C3 ways for this year and another group of 2 cities for next year in 7C2 ways. That will give me possible 10C3 * 7C2 of visiting 5 cities.
Here {(A, B, C), (D, E)} is different from {(A, B, D), (C, E)} but in our original question, both are one selection only.

rsrighosh

KarishmaB

Great explanation. Thanks!

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