Four contestants representing four different countries advance to the

As I see we talk about how many possibilities of placing 4 distinct objects
with regard to 1 and 2 place

All possibilities: 4!=24

Restrictions: two places regard and two not regard multiplyed by 2 so, (4!/2!*2!)*2=12


C

The correct answer is C.

isnt 4C2 (i.e 6 ways) the answer ?
please clarify

Thanks

No. 4C2 gives the number of two-contestant groups possible from 4. Suppose we are choosing group {A, B}. We can have A = first-place and B = second-place or vise-versa, so for each group there are 2 possible arrangement with respect to how a first-place and second-place medal can be awarded. Therefore the answer is 4C2*2 = 12.

Answer: C.

Hope it's clear.

Yes, thank you
another way to look at it is : 4C1 * 3C1 = 12 possible ways.

Hi,

Let me share the 2 ways in which I did it. Please, let me know if there is a mistake somewhere in my thinking or a possible trap in these 2 solutions:

1st solution:
4!=24 possible ways
24/2=12, because it is the first 2 spots that will be changing. It is sort of intuitive, so it could easily be wrong...

I tried the 2nd solution solution, just to see if I end up with 12 again:
We have A - B - C - D as the possible slots. The first 2 slots can change so that it is occupied by 2 different people.
Starting with A we have:
A - B - C - D
A - C - B - D
A - D - B - C
So, these are 3 ways, only taking into account A. We gave 4 different people, so 3 X 4 = 12 (just thinking that each one of them should have the same chances).

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