Question

Among 5 students how many ways can we give 1st and 2nd pricec

Answer 1

Ans: 720

Solution 1

Required number of ways is equal to number of arrangements of 3 from 10 distinct objects and hence answer is 10P3 = 720

Answer 2

I'm not sure about this answer but any I'm trying to answer it

we can do this by the principle of permutations

nPr

=5P2

=5!/(5-2)!

=5!/3!

=5×4×3!/(3!)

=5×4

=20

therefore we can give the first and second price for 5 students in 20 different ways