The number of ways in which 15 identical apples and 10 identical oranges

The number of ways in which 15 identical apples and 10 identical oranges

since 15 apples are identical, let us place the 15 apples in a row leaving a space before, after and between them _a_a_a_a_a_a_a_a_a_a_a_a_a_a_a_ place a divider in any of the 16 spaces and place another divider in any of 16 spaces.

this can be done in C216=16*152=120 ways

there are 16 ways to place both the divider on the same place .

_a_a_a |a_a_a_a_a_a | a_a_a_a_a_a_ represents (3,6,6)_a_a_a _a_a_a_a_a_a || a_a_a_a_a_a_ represents (9,0,6)

therefore total number of ways to distribute 15 apples among three persons = 120+16 = 136 ways similary 10 identical oranges can be distributed among three persons in

C211+11=11*102+11

= 55+11 = 66 ways therefore here are 66 ways to distribute 10 identical oranges among three persons. now total number of ways  = 136*66 = 8976 ways

hope this helps you

You have to distinguish between the different types of fruits.

There are three possible recipients for the orange, three possible recipients for the plum, and three possible recipients for the tangerine. As for the six apples, let $x_i$ be the number of apples received by person $i$, $1 \leq i \leq 3$. Then $$x_1 + x_2 + x_3 = 6$$ is an equation in the nonnegative integers. A particular solution of the equation corresponds to the placement of $3 - 1 = 2$ addition signs in a row of $6$ ones. For instance, $$1 1 + 1 1 + 1 1$$ corresponds to the solution $x_1 = x_2 = x_3 = 2$, while $$1 1 1 1 + 1 1 +$$ corresponds to the solution $x_1 = 4, x_2 = 2, x_3 = 0$. The number of such solutions is the number of ways we can place two addition signs in a row of six ones, which is $$\binom{6 + 3 - 1}{3 - 1} = \binom{8}{2}$$ since we must select which two of the eight positions required for six ones and two addition signs will be filled with addition signs.

Hence, the number of ways six indistinguishable apples, one orange, one plum, and one tangerine can be distributed to three people without cutting any fruit is $$\binom{8}{2}3^3$$

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The number of ways of distributing 20 identical fruits among 5 people, so that no one receives less than 3 fruits is

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The number of ways in which 15 identical apples & 10 identical oranges can be distributed among three persons , each receiving none,one or more is 

No. of ways in which $$15$$ identical apples are distributed among three person $$=^{ 15+3-1 }{ { C }_{ 3-1 } }=^{ 17 }{ { C }_{ 2 } }=136$$

No. of ways in which 10 identical oranges are distributed among three persons $$=^{ 10+3-1 }{ { C }_{ 3-1 } }=^{ 12 }{ { C }_{ 2 } }=66$$

Total no. of ways $$136\times 66=8976.$$

Hence, the answer is $$8976.$$