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nivas1811

  • 3 years ago

in how many ways can 5 prizes be given away to 3 boys when each boy is eligible for one or more prizes..?

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  1. nicolebears
    • 3 years ago
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    |dw:1352712893295:dw|

  2. mayankdevnani
    • 3 years ago
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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.

  3. mayankdevnani
    • 3 years ago
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    ok @nivas1811

  4. Aperogalics
    • 3 years ago
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    @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:)

  5. Aperogalics
    • 3 years ago
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    @nivas1811

  6. nivas1811
    • 3 years ago
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    @mayankdevnani,@Aperogalics : there are five prizes so there are 3^5 ways to distribute the prizes...

  7. Aperogalics
    • 3 years ago
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    @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 :)

  8. Aperogalics
    • 3 years ago
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    and if one prize is given away then how same can be given to other:)

  9. shubhamsrg
    • 3 years ago
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    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

  10. shubhamsrg
    • 3 years ago
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    formulla is for x1 + x2 + x3 .... xr = n integral solutions is (n+r-1 , r-1) given that each x1,x2... is >=0

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