Among 5 students how many ways can we give 1st and 2nd pricec
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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
Solution 1
Required number of ways is equal to number of arrangements of 3 from 10 distinct objects and hence answer is 10P3 = 720
Answered by
0
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
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
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