solve by generating function along with approach thanks advance

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  1 or more than 1 fruits.

Comments

okay

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hacker16 I think your approach has a problem. by considering all fruits together and total 9 fruits, we are trying to see all fruits as alike. in fact we are trying here to solve the problem of dividing 9 identical fruits to 3 baskets so that each basket gets at least one ($x_1+x_2+x_3 = 9 | x_1\geq 1 ,x2 \geq 1,x_3 \geq 1$). here we are not considering some of the distinct cases and treating them all as alike. 

for e.g, we assume there are total 9 fruits. then in your way, one combination is 2 3 4. but it is not single combination as we can have multiple combinations for this single case. such as, 

1) 1A1M   1A1M1O  2O2M

2) 2O        2A1M    3M1O

your approach took all these distinct cases into one single case. 

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thanks for pointing out my mistake @thanks_junk_mayavi

Answer 1

Way to approach :- First distribute 2 identical apples,3 identical fruits and 4 identical mangoes to those 3 children without taking care about any bound about at least how much every child will eat. Then out of those ways subtract those ways When any one child out of three will have $0$ fruits and other will have non-zero fruits and then out of those remaining ways subtract those ways when any two children out of three can have $0$ fruits and third child will have non-zero fruit.

Part 1) Distribute those fruits without any bound

a) Distribute $2$ Apples

$C_{1}+C_{2}+C_{3}=2$   

Total ways $=$$\binom{4}{2}$ ways

b) Distribute $3$ Oranges

$C_{1}+C_{2}+C_{3}=3$   

Total ways $=$$\binom{5}{3}$ ways

C) Distribute $4$ Mangoes

$C_{1}+C_{2}+C_{3}=4$   

Total ways $=$$\binom{6}{4}$ ways

So overall Total ways to distribute $2$ Apples , $3$ Oranges , $4$ Mangoes to $3$ children , when any one can have any number of fruits $=$ $\binom{4}{2}*\binom{5}{3}*\binom{6}{4}$ $=900$ ways

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Part 2) When one child out of three will have $0$ fruit and then distribute fruits in similar way as above.

Choose 1 child out of three , that will have $0$ fruit = $\binom{3}{1}$ ways

Let suppose $C_{1}=0$

So now again three parts

a) $C_{2}+C_{3}=2$ 

$\binom{3}{2}$ ways

b) $C_{2}+C_{3}=3$ 

$\binom{4}{3}$ ways

c) $C_{2}+C_{3}=4$ 

$\binom{5}{4}$ ways

So total ways when $C_{1}$ has no fruit = $3*4*5=60$ , now we can see out of those ways there is case when $C_{2}$ receive no fruit and all fruits are distributed to $C_{3}$ and similar when all fruits are distributed to $C_{2}$ so subtract those two ways as , we are only considering those cases in that part when all fruits are distributed to $C_{2}$ and $C_{3}$ and each have at least one fruit

So overall of part 2) = $\binom{3}{1}$$*(60-2)$ $=174$

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Part 3) When all fruits are distributed to only one child , simply one way and to choose that child , 3 ways so overall = 3 ways

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So answer = part 1 - (part 2+part 3)

Hope i didn't make any mistake , verify the answer.

Comments

There are some redundant subtractions in part 2, when you subtracted 2 from 60.

Let's say C1 was initially selected for getting 0 fruits. Then we distributed all fruits among C2 and C3, this gave 60.

Now, we subtracted 2 cases, these are (i)when C2 get 0 and (ii)when C3 get 0.

Look at (i) This is the case when C1 and C2 get 0.

When C2 is initially selected for getting 0. Distribute among C1 and C3 gives 60 ways.

Subtracting two cases from it: (i)when C1 get 0 and (ii)when C3 get 0.

Here (i) is C1 and C2 get 0.

This was already subtracted.

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Hello Rishbh

Yes! same thing subtracted multiple times because same thing was added multiple times.

Look when $C_{1}=0$ we're distributing fruits b/w $C_{2}$ and $C_{3}$ and in that way there is possibility that one of them also got $0$ , which we don't want to here we subtracted those cases

Now when $C_{2}=0$ and we're distributing fruits b/w $C_{1}$ and $C_{3}$ , there is a possibility that again one receive $0$ and other received all , so again we have to subtract those cases.

Look this venn diagram , when we need to find section like "only $A$ or only $B$" From $A$ we have to subtract $A \cap B$ and from $B$ also we have to subtract $A \cap B$. Multiple times it was added so multiple times subtracted.

Answer is right.I have rechecked it today.

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yes answer 723 i also verified

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Ok I got it.
Thanks Rupendra :)

Can you tell how to do this using generating function.?