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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
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Any one prize can be given to any one of the 3 boys and hence there are 3 ways of distributing each prize.
Hence, the 4 prizes can be distributed in 3^4= 81 ways.
@mayankdevnani
there are 3 boys so first boy can have 1,2 or 3 prizes
now for 3,1,1 we have 3ways-(3,1,1),(1,3,1),(1,1,3)
now for 2,2,1 we have 3 ways-(2,1,2),(2,2,1),(1,2,2)
so only 6 ways to distribute:)
@nivas1811
sry in 3^5 a case comes where 5 prizes are given to one person but you wrote that
each boy is eligible for one or more prizes.. so no prize case is eliminated :)
let 1st guy get x prizes, 2nd one y and third one z
we are given x+y+z = 5
where
x>=1
y>=1
z>=1
we can write the eqn as
(x-1) + (y-1) + (z-1) = 2
where now each term inside bracket >=0
it has a direct formulla now for integral solutions of the eqn,,who's background isnt much complex and i can tell you if you desire..
ans would be
4C2 or 6