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In how many ways can we give 5 apples, 4 mangoes, 3 oranges to 3(fruits of the same species are similar people if each may receive none, one or more.

I have use the stars and bars method:
We have to find the no. of possible arrangements of 5 apples, 4 mangoes, 3 oranges and 2 partitions. This can be given by $\frac{14!}{5!4!3!2!}$. But the answer is supposed to be 3150 and I can't figure out why.

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You have to distribute each type of fruit separately.

Let $a_k$, $1 \leq k \leq 3$, be the number of apples distributed to the $k$th person. Then $$a_1 + a_2 + a_3 = 5 \tag{1}$$ The number of ways the apples can be distributed is the number of solutions of equation 1 in the nonnegative integers. A particular solution corresponds to the placement of two addition signs in a row of five ones. For instance, $$1 1 + 1 + 1 1$$ corresponds to the solution $a_1 = 2$, $a_2 = 1$, and $a_3 = 2$, while $$+ 1 1 1 + 1 1$$ corresponds to the solution $a_1 = 0$, $a_2 = 3$, and $a_3 = 2$. Therefore, the number of such solutions is the number of ways two addition signs can be inserted into a row of five ones, which is $$\binom{5 + 2}{2} = \binom{7}{2}$$ since we must choose which two of the seven symbols (five ones and two addition signs) will be addition signs.

The number of ways the mangoes can be distributed is the number of solutions in the nonnegative integers of the equation $$m_1 + m_2 + m_3 = 4 \tag{2}$$ where $m_k$, $1 \leq k \leq 3$, is the number of mangoes distributed to the $k$th person. The number of such solutions is

$$\binom{4 + 2}{2} = \binom{6}{2}$$

The number of ways the oranges can be distributed is the number of solutions in the nonnegative integers of the equation $$o_1 + o_2 + o_3 = 3 \tag{3}$$ where $o_k$, $1 \leq k \leq 3$, is the number of oranges distributed to the $k$th person. The number of such solutions is

$$\binom{3 + 2}{2} = \binom{5}{2}$$

To find the number of ways of distributing the three types of fruit, multiply the above results.

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    $\begingroup$ But why does my answer give a wrong result...I mean whats wrong with that logic? $\endgroup$ – Osheen Sachdev Feb 26 '17 at 2:41
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    $\begingroup$ Apples are distinguishable from mangoes, mangoes are distinguishable from oranges, and apples are distinguishable from oranges. Therefore, each fruit must be treated separately. $\endgroup$ – N. F. Taussig Feb 26 '17 at 2:46
  • $\begingroup$ Your calculation $$\frac{14!}{5!4!3!2!}$$ represents the number of distinguishable arrangements of five apples, four mangoes, three oranges, and two partition points or, equivalently, distinguishable permutations of the letters of the word AAAAAMMMMOOOPP. However, the order in which each person receives the pieces of fruit does not matter. That is why I treated the problem as three combinations with repetition problems. Your answer is too large since you treated a person who receives an apple, a mango, and an apple in that order differently than one who receives a mango then two apples. $\endgroup$ – N. F. Taussig Feb 26 '17 at 6:46

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