# Easy Permutation-Combination Solved QuestionAptitude Discussion

 Q. There are three prizes to be distributed among five students. If no students gets more than one prize, then this can be done in:
 ✔ A. $10$ ways ✖ B. $30$ ways ✖ C. $60$ ways ✖ D. $80$ ways

Solution:
Option(A) is correct

$3$ prizes among $5$ students can be distributed in $^5C_3$ ways $= 10$ ways.

Edit: Thank you Rakesh for explaining why ans is not $5 \times 4 \times 3 = 60$ and the assumption behind this solution.

## (8) Comment(s)

Ankit
()

let 5 students be A B C D E , if one gets the prize mark them as 1 , if not mark them as 0. then find the no. of arrangement of 5 bit binary pattern. if there are 3 one's and 2 are zeroes's.

ABHIJEET
()

5 places and 3 things to arrange? assume it like this XXX00. X= prizes (identical) 0 = Nothing.. so if u have learnt arranging identical items..u'll know arranging XXX00 is 5!/(3! x 2!) = 10 :)

ANUPAM KUMAR TIWARI
()

Why the answer is not 60

As prize1 can be given in 5 ways prize2 in 4 ways prize3 in 3 ways so overall 60 ways

Rakesh
()

While solving it is assumed that prizes are identical. Thus the problem reduces to choosing 3 students among 5 students (and there is only 1 way of distributing, since prizes are identical).

Think about it and you will understand why it is true.

Had the prizes been different, there would have been $3!$ ways of distributing it among the three students. Making total ways to be:

$^5C_3 \times 3! = 10 \times 3 \times 2$

$=\textbf{ 60 ways}$

PRATYUSH ANAND
()

As all the prizes are same . So if any 3 students get these prizes no need to find the way he can get other prize.

So if we consider 5 students as A,B,C,D,E and F.

Then 3 prizes would be given to-->

A B C ---1 way

or

A B D ----2 ways

or

A B E ----3 ways

or

A B F ----4 ways

or

B C D ----5 ways

or

B C E ----6 ways

or

B C F ----7 ways

or

C D E ----8 ways

or

C D F ----9 ways

or

D E F ----10 ways

NOTE:- We should not consider A B C as one way and B A C as other way, because prizes are same.

As here no. of boys to select from 5 i.e 3 is equal to no. of prizes i.e 3

and we have to select not arrange after getting the prize so Combination will be used.

By formula Total ways, $^5C_3=10$ ways.

Venki
()

who said that the prizes are equal.........they may be 1st,2nd and 3rd prizes of running competetion.........i can assume like this also........so 60 should be correct from my perspective.........is there any notice that if nothing mentioned assume all are identical........????

Raj Singh
()

Instead of Prizes, had it been Apples, we can consider them to be identical. In fact, question should be re-phrased because Prizes, in general, cannot be considered as identical.

VeeraragavanK
()

One prize can be given 5 ways. Next, 2nd prize is given in 4 ways to any of the 4 students. 3rd prize can be given in 3 ways.

In all, $5+4+3= 12$ ways,Again, 4 prizes can be given amongst the 3 students in 3! ways.

Hence, total number of ways $=12*6=72$?