Bunuel wrote:
pappueshwar wrote:
hey all,
whats the answer?
i got it as 60 . is that wrong ?
OA is given under the spoiler in the initial post.
Five people are running in a race. The first one to finish wins a gold medal, the second wins a silver medal and the third wins a bronze medal. How many different arrangements of medal winners, in order from first to third, are possible?A. 5
B. 10
C. 60
D. 120
E. 125
\(C^3_5*3!=60\), where \(C^3_5\) is ways to select the group of 3 winners out of 5 contestants and 3! is ways to arranging them (alternately you can just do \(P^3_5=60\)).
Answer: C.
Hope it helps.
Hello Bunuel,
I didn't understand why you used \(P^3_5=60\) to solve this question. Also, the actual evaluation of it seems to get me - I know this is a silly point but I still get it wrong -
I evaluate as follows:
\(P^3_5\)= 5!/3! = 5*4*3*2*1/ 3*2*1 = 20. where am I going wrong?
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