In how many ways can three people divide amongst themselves six identical apples, one orange,one plum and one tangerine(without cutting any fruit)?
Well, my solution goes something like this:
We have $9$ fruits in total to be distributed amongst $6$ persons.
Now, we can represent this as:$x_1+…+x_6=9$ . The number of solutions of the equation is $15\choose 5$. So, the $9$ fruits can be distributed among those $6$ persons in $ 15\choose 5$ ways.
However, the answer is given as $8\choose 2$$3.$$3.$$3$ ways. How is this is possible? Where is the problem occuring? I am not getting it…
Best Answer
You have to distinguish between the different types of fruits.
There are three possible recipients for the orange, three possible recipients for the plum, and three possible recipients for the tangerine. As for the six apples, let $x_i$ be the number of apples received by person $i$, $1 \leq i \leq 3$. Then $$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$$