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Combination

by sam2304 » Fri Jun 01, 2012 6:49 pm
A Â…firm is divided into four departments, each of which contains four people. If a project is to be assigned to a team of three people, none of which can be from the same department, what is the greatest number of distinct teams to which the project could be assigned?
(A) 4^3
(B) 4^4
(C) 4^5
(D) 6(4^4)
(E) 4(3^6)
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by eagleeye » Fri Jun 01, 2012 7:04 pm
Hi sam2304:

We have 4 departments with 4 people in each. To select 3 members, each one from a distinct team, we need to select 1 person each from 3 of the teams.

Step 1: Choose 3 teams out of 4. No of ways = 4C3 = 4
Step 2: Choose 1 person each from each of the selected teams. No of ways = 4C1*4C1*4C1 = 4*4*4
Total ways = (Choose 3 teams out of 4)*(select one person each of the 3 teams) = [spoiler]4*4*4*4 = 4^4. Hence B[/spoiler].

Another way: (By permutation and then combination)

There are total 16 people. We can select the first person from 16, then the second from 12, and the third from remaining 8. (Since we can only select 1 from each department).
So, no of ways = 16*12*8. Now the order of selection of those people is not important. Hence, we divide by 3!

Total no. of ways = (16*12*8)/3!= [spoiler]4^4. Hence still B[/spoiler]

Let me know if this helps :)
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by sam2304 » Fri Jun 01, 2012 11:42 pm
OA : B I solved using the first method posted by you, but wasn't sure whether it was the right approach. The doubt is cleared now :)
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