in how many ways can 5 prizes be given away to 3 boys when each boy is eligible for one or more prizes..?
Stacey Warren - Expert brainly.com
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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.
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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:)
: there are five prizes so there are 3^5 ways to distribute the prizes...
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 :)
and if one prize is given away then how same can be given to other:)
let 1st guy get x prizes, 2nd one y and third one z
we are given x+y+z = 5
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
x1 + x2 + x3 .... xr = n
integral solutions is (n+r-1 , r-1)
given that each x1,x2... is >=0