Permutation and combination

Question

in how many ways 2 alike apple, 3 alike orange and 4 alike mango can be given to 3 children if each child can have none or 1 or more than 1 fruits.

Comments

900. ways.

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https://math.stackexchange.com/questions/2161696/in-how-many-ways-can-we-give-5-apples-4-mangoes-3-oranges-to-3-people-if-each-ma

Answer 1

n =3  childrens

x1 + x2 + x3 = 2   //apple

r = 2

ways =  $_{r}^{n+r-1}\textrm{C}$ = $_{2}^{4}\textrm{C}$  =  6

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x1+x2+x3 =  3  // orange   

r = 3

ways = $_{r}^{n+r-1}\textrm{C}$ = $_{2}^{5}\textrm{C}$   = 10

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x1+x2+x3 = 4 // mango

r = 4

 ways  =  $_{r}^{n+r-1}\textrm{C}$  =  $_{4}^{6}\textrm{C}$ = 15

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total ways = $6\times 10\times 15 = 900$

Thank you !

Comments

but what if i change language to each children can receive  1 or more than fruits , then how to solve , means a child  cannot get zero fruits  here , if anyone knows share approach

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Use geneartaing function , 49 ways

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Anu007   provide approach not good with generating function  , share  image of solved  answer 

Thank you 

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or link of alike question where generating function used ,will be helpful