Latest IBM Aptitude Question SOLUTION: How many ways can 4 prizes be given away to 3 boys, if each boy is eligible for all the prizes?

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How many ways can 4 prizes be given away to 3 boys, if each boy is eligible for all the prizes?

Asked In IBM (13 years ago)

Unsolved Read Solution (10)

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Read Solution (10) :

princi  (13 years ago)

The 4 prizes can be distributed to 3 boys in 3^4 =81ways

Is this solution Helpfull? Yes (38) | No (19)  

Phani Kiran  (13 years ago)

4prizes can be distributed to 3 boys in 4P3 ways(permutation) i.e.,4x3x2=24ways

Is this solution Helpfull? Yes (35) | No (13)  

Poulami Ghosh  (12 years ago)

15 ways.
b1 b2 b3
no of prize 4 0 0
3 1 0
3 0 1
2 2 0
2 0 2
2 1 1
1 3 0
1 0 3
1 2 1
1 1 2
0 4 0
0 0 4
0 3 1
0 1 3
0 2 2

Is this solution Helpfull? Yes (7) | No (9)  

Rahul kumar singh  (12 years ago)

For the first time aprize can be any one of the 4. once a prize is given away remaining 3 prizes can be given away in 3c2 ways to the remaining two people.So each prize has 3c2 i.e. 6 ways of giving associated with it .So for 4 prizes total no of ways =6*6*6*6=6^4 ways.

Is this solution Helpfull? Yes (3) | No (7)  

Veluru Sathyavathi  (8 years ago)

Let the 3 boys be B1, B2, B3 and 4 prizes be P1, P2, P3 and P4.

Now B1 is eligible to receive any of the 4 available prizes (so 4 ways)
B2 will receive prize from rest 3 available prizes(so 3 ways)
B3 will receive his prize from the rest 2 prizes available(so 2 ways)

Finally B4 will receive the the last remaining prize (so 1 way)
So total ways would be: 4*3*2*1=24 Ways

Hence, the 4 prizes can be distributed in 24 ways.

Is this solution Helpfull? Yes (2) | No (1)  

ASHOKCHAKRAVARTHI N  (13 years ago)

first boy would get one prize from 4 available prizes,
second boy would get one prize from remaining 3 available prizes,
third boy would get one prize from remaining 2 available prizes,
So 4*3*2=36

Is this solution Helpfull? Yes (1) | No (20)  

Chandan Kumar  (10 years ago)

(n+r-1)C(r-1)=6C4= 15.

Is this solution Helpfull? Yes (1) | No (1)  

sonali Akkole  (8 years ago)

correct answer?

Is this solution Helpfull? Yes | No   

ANANYA BIHARI  (5 years ago)

Each prize can be given to 3 boys

no of ways= 3^4=81 ways(n^r where n=3 and r=4)

Is this solution Helpfull? Yes | No   

Ramya Easwaran  (5 years ago)

Let the 3 boys be B1, B2, B3 and 4 prizes be P1, P2, P3 and P4
Now B1 is eligible to receive any of the 4 available prizes (so 4 ways)
B2 will receive prize from rest 3 available prizes(so 3 ways)
B3 will receive his prize from the rest 2 prizes available(so 2 ways)
So total ways would be: 4 × 3 × 2 × 1 = 24 Ways
Hence, the 4 prizes can be distributed in 24 ways

Is this solution Helpfull? Yes | No   

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