Adi_Pat wrote:From a group of 21 people - out of whom 12 have experience - a 3 person crew is selected such that only 1 has experience. How many different crews can be selected ?

1) 432
2) 594
3) 864
4) 1330
5) 7980

OA A

I am usually not great with Combination problems, but I'll try explaining this one.

We need to select 1 person out of 12 experienced ppl. So 12C1 ways to select = 12 ways to select that 1 person.
We need to select 2 people out of 9 non-experienced ppl. So 9C2 ways to select = 36 ways to select those 2 people.
Total ways = 12 * 36 = 432. (A)

Makes sense with the formula...Why is this approach giving me the wrong answer...

Number of ways of selecting 1 experienced person = 12

So that leaves 9 people....

Number of ways of selecting 2nd person = 9
Number of ways of selecting 3rd person = 8

Total number of ways = 12*9*8 = 864

Adi_Pat wrote:Makes sense with the formula...Why is this approach giving me the wrong answer...

Number of ways of selecting 1 experienced person = 12

So that leaves 9 people....

Number of ways of selecting 2nd person = 9
Number of ways of selecting 3rd person = 8

Total number of ways = 12*9*8 = 864

Nowhere in the question does it specify that the 2 inexperienced people are picked one by one, which is I believe what your approach is assuming.

Adi_Pat wrote:Makes sense with the formula...Why is this approach giving me the wrong answer...

Number of ways of selecting 1 experienced person = 12

So that leaves 9 people....

Number of ways of selecting 2nd person = 9
Number of ways of selecting 3rd person = 8

Total number of ways = 12*9*8 = 864

Hi! You've worked out the problem as though order matters, which is why you got 72 instead of 36 for the second step (inexperienced) part of the question.

If the question had read:

3 seats in a canoe are going to be filled by 1 experienced person and 2 inexperienced people. There are 12 experienced canoeists and 9 inexperienced canoeists available. If the experienced canoeist must sit in the back and the inexperienced canoeists will occupy the first two seats, in how many ways can the 3 seats be filled?

then your solution would have been perfect. However, in the actual question we don't care about order, so you've double counted.

Let's call the inexperienced people A, B, C, D, E, F, G, H and I.

The way you've solved the problem, you could select A first and then B second; however, you could also select B first and then A second. You've counted these as 2 distinct selections, even though they're both simply A + B selected.

So, you need to divide by 2 to eliminate the double counting.

12 * (9 * 8)/2 = 12 * 72/2 = 12 * 36 = 432