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$4$ identical oranges and $6$ distinct apples(each of different variety) are to be distributed into $5$ distinct boxes. Then probability that each box get a total of $2$ objects is

Try: Number of ways in which $4$ identical object of one type and $6$ Distinct objects of other type are to be distributed into $5$ persons such that each have $2$ objects is

$$=\binom{5}{2}\times 1 \times \binom{6}{2}\times \binom{4}{2}\times \binom{2}{2}$$

Now i did not understand how to solve it, Help me to solve it , thanks

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    $\begingroup$ How are the objects distributed? Can someone get 0 objects? $\endgroup$ – Joseph Eck May 2 '18 at 3:20
  • $\begingroup$ Please elaborate more about how you distribute those fruits. $\endgroup$ – Postal Model May 2 '18 at 4:00
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Number of distributions of four identical oranges and six distinct apples to five distinct boxes: Since the oranges are identical, what matters is how many are placed in each box. Let $x_k$ be the number of oranges that are placed in the $k$th box. Then
$$x_1 + x_2 + x_3 + x_4 + x_5 = 4$$ is an equation in the nonnegative integers. A particular solution of the equation corresponds to the placement of four addition signs in a row of four ones. For instance, $$+ 1 1 + 1 + + 1$$ corresponds to the solution $x_1 = 0$, $x_2 = 2$, $x_3 = 1$, $x_4 = 0$, $x_5 = 1$. The number of such solutions is $$\binom{8}{4}$$ since we must select which four of the eight positions required for four ones and four addition signs will be filled with addition signs.

Since the apples are distinct, what matters is which apple is placed in which box. There are five distinct boxes in which each of the six distinct apples could be placed. Thus, there are $5^6$ ways to distribute the apples.

Thus, there are $$\binom{8}{4}5^6$$ ways to distribute four identical oranges and six distinct apples to five distinct boxes.

Number of distributions of four identical oranges and six apples to five distinct boxes in which two pieces of fruit are placed in each box: We place the oranges, then distribute the apples.

Two oranges are placed in each of two boxes and two apples are placed in each of the other three boxes: The two boxes that each receive two oranges can be selected in $\binom{5}{2}$ ways. Of the remaining three boxes, there are $\binom{6}{2}$ ways to select which two apples are placed in the lowest numbered of those boxes and $\binom{4}{2}$ ways to select which two of the remaining four apples are placed in the next lowest numbered of those boxes. The remaining two apples must be placed in the other box. There are
$$\binom{5}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}$$ such distributions.

Two oranges are placed in one box, one orange and one apple are placed in each of two other boxes, and the other two boxes each receive two apples: There are five ways to select the box that receives two oranges and $\binom{4}{2}$ ways to select which two of the four remaining boxes receive one orange each. There are six ways to place an apple in the lower numbered of the two boxes with one orange and five ways to place an apple in the higher numbered of those boxes. There are $\binom{4}{2}$ ways to place two of the four remaining apples in the lower numbered of the two remaining boxes. The other box must receive the other two apples. There are $$\binom{5}{1}\binom{4}{2}\binom{6}{1}\binom{5}{1}\binom{4}{2}\binom{2}{2}$$ such distributions.

One orange and one apple are placed in each of four boxes and the remaining two apples are placed in the remaining box: There are $\binom{5}{4}$ ways to select which boxes receive an orange. There are $\binom{6}{2}$ ways to select which two apples are placed in the empty box. The remaining four apples can be distributed to the four boxes that contain an orange so that one apple is placed in each such box in $4!$ ways. Hence, there are $$\binom{5}{4}\binom{6}{2}4!$$ such distributions.

Total: The number of ways of distributing four identical oranges and six distinct apples to five distinct boxes in such a way that each box receives two pieces of fruit is $$\binom{5}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2} + \binom{5}{1}\binom{4}{2}\binom{6}{1}\binom{5}{1}\binom{4}{2}\binom{2}{2} + \binom{5}{4}\binom{6}{2}4!$$

Probability that four identical oranges and six distinct apples are distributed to five distinct boxes in such a way that two pieces of fruit are placed in each box: Divide the number of ways of distributing the pieces of fruit in such a way that two pieces of fruit are placed in each box by the number of ways of distributing the fruit.

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Since I don't know how you distribute those objects I'll just solve the case each have two fruits.

First we put fruits into five bags, each bag containing two fruits. Next consider the distribution of those identical oranges.

$(2,2,0,0,0)$: $$\textrm{(choose two lucky person each get a bag of two oranges)}\cdot(\textrm{the others get apples}).\\ {5\choose2}\cdot{6\choose2}{4\choose2}{2\choose2},$$

$10\cdot15\cdot6=900.$

$(2,1,1,0,0)$: $$\textrm{(choose one lucky guy get the only bag of two orangs)}\cdot(\textrm{choose two apples to pair the two oranges}\cdot\textrm{choose two guys to receive apple-orange pair})\cdot(\textrm{distributing apple-apple bags}).\\{5\choose1}\cdot({6\choose2}\cdot{4\choose2}2!)\cdot{4\choose2}{2\choose2},$$

$5\cdot(15\cdot12)\cdot6=5400.$

$(1,1,1,1,0)$: $$(\textrm{choose two apples which won't pair with an orange})\cdot(\textrm{choose one guy receive this apple-apple bag})\cdot(\textrm{distribute four apple-orange pairs})\\ {6\choose2}\cdot{5\choose1}\cdot4!,$$

$15\cdot5\cdot24=1800.$

So $\Large\frac{1800+5400+900}{?}$, which is depending on whether or not you're allowing a person receive nothing.

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  • $\begingroup$ In your calculation of the case that two lucky people each get a bag of two oranges and the others get apples, you should have obtained $10 \cdot 15 \cdot 6 = 900$. Everything else is correct. $\endgroup$ – N. F. Taussig May 3 '18 at 10:23

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