Distribution Of Things - Equation and Solved Examples

Distribution of things is used to find the number of ways of distributing n different things in r different boxes. When identical objects are distributed among persons, the only thing that matters is who is getting how many objects. In real life, we use the distribution of things to distribute different things in different boxes.

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In this article, we will learn about the Distribution Of Things. This topic falls under the broader category of Permutations and combinations, which is a crucial chapter in Class 11 Mathematics. This is very important not only for board exams but also for competitive exams, which even include the Joint Entrance Examination Main and other entrance exams: SRM Joint Engineering Entrance, BITSAT, WBJEE, and BCECE.

Distribution Of Things: definition

Distribution of things is used to find the number of ways of distributing n different things in r different boxes. Firstly, we will examine the distribution of distinct items in distinct boxes where empty boxes are permissible and then we will investigate the distribution of distinct items in distinct boxes, but in this case, empty boxes are not allowed. Lastly, we will explore the distribution of identical items in distinct boxes where empty boxes are permitted

Conditions for Distributions of Things

There are two conditions for the distribution of things

1) No restrictions

2) Non-empty restrictions

No restriction

To distribute n different things in r different boxes, such that there is no restriction on the number of objects a box can have (some boxes can remain empty as well)

We have r options for each object, so the number of ways is r.r.r......n times = rⁿ

Non-empty restriction

To distribute n different things in r different boxes, such that none of the boxes is empty

Such problems can be resolved in the following steps

1. Decide the number of objects in each group and make cases accordingly

2. In each case, first divide them into groups using the grouping formula, then distribute these r groups into r people

Suppose we need to distribute 5 different hats in 3 different boxes such that no box is empty

Step 1:

To decide the group sizes, we should first distribute 1 hat to each 3 boxes so that none of them remain empty. Now we will be left with 2 hats, those 2 hats can be distributed to

1 group forming group combination as 1 1 3

OR

1-1 hat to 2 different groups giving the combination 1 2 2.

So in that way, we will get two cases

Step 2.

Case I:
Make a group of sizes $(113)$ combination in $\frac{5!}{1!1!3!} \cdot \frac{1}{2!}$ ways.
Now distribute these 3 groups in 3 people in 3 ! ways
So the total number of ways for case $1=\frac{5!}{1!1!3!} \cdot \frac{1}{2!} \cdot 3!=60$

Case II:
Following the same logic in this case
Total number of possible ways of distribution $\frac{5!}{1!2!2!} \cdot \frac{1}{2!} \cdot 3!=90$

Hence total ways = 60+90 = 150

This type of concept can be comprehended as arranging all n identical objects to be distributed and (r-1) marks of partition. These (r-1) partitions will divide the objects in r groups. These r groups can be given to these r distinct people in order.

As each person can get zero or more objects in such an arrangement, so number of ways is

$
\frac{(n+r-1)!}{(r-1)!n!}={ }^{n+r-1} C_{r-1}
$

If each one has to get at least one object, then first distribute r objects to these r people (each one gets one, and any r objects can be given as these are identical), then we can distribute the remaining ( $\mathrm{n}-\mathrm{r}$ ) objects in r people in ${ }^{n-r+r-1} C_{r-1}={ }^{n-1} C_{r-1}$ ways

Alternative Method

This thing can also be comprehended as the following

Let the first group get a₁ objects

The second group gets a₂ objects

.....

r^th group gets a_r objects

So, a₁ + a₂ + a₃+...+ a_r = n …....(i)

Now if empty groups are allowed we need to find the whole number solution of the equation (i)

So, whole number solutions of $a_1+a_2+a_3+\ldots+a_r=n$ equal the number of ways to distribute $n$ distinct objects in r people, and these both equal ${ }^{n+r-1} C_{r-1}$

If empty groups are not allowed, then each of these values has to be a natural number. The number of ways of distribution will thus equal the natural number of solutions of equation (i), which is ${ }^{n-1} C_{r-1}$

Example: In how many ways can 10 Identical chocolates be distributed among 3 children, such that each student can get any number and at least one?

Solution: using the above concept, we use the direct formula for this, so we have

$
{ }^{10-1} C_{3-1}={ }^9 C_2
$

Suppose we want to distribute 5 distinct hats in 3 Identical boxes such that each box receives at least 1 hat. This can be done in 2 steps

Step 1: Decide the number of hats that each will get. Make cases depending on this

Step 2: Make groups in each case using the formula for grouping

Now,

Step 1: we can distribute 3 hats one-one each to all 3 groups, after that, we can place the remaining 2 hats in one group or 1-1 to 2 groups. So 2 cases: 1^st = 1 1 3, 2^nd = 1 2 2

Step 2: Now these cases are similar to the division of groups where group sizes are given

So, according to $1^{\text {st }}$ case, we can be group hats in $\frac{5!}{(1!)^2 3!} \cdot \frac{1}{2!}$ ways
Similarly for $2^{\text {nd }}$ case, $\frac{5!}{(2!)^2 1!} \cdot \frac{1}{2!}$ ways
So the total number of ways of distribution $=\frac{5!}{(2!)^2 1!} \cdot \frac{1}{2!}+\frac{5!}{(1!)^2 3!} \cdot \frac{1}{2!}$ ways

In this type, it does not matter which object goes in which group as all objects are identical, the only thing that matters is how many objects go into groups, and that means ordering or group does not matter.

Example: In how many ways can 12 identical hats be put in 3 identical boxes such that each box has at least 2 hats?

Solution: First and foremost 2 hats should be put in each box (which hats do not matter as all are identical). Now we are left with 6 hats to be put in 3 identical boxes. As learned earlier, in the case of identical boxes, we are only concerned with dividing the 6 hats into 3 portions.

The hurdle is that the sizes could be anything. Thus again we individually consider all cases, with groups being every possible size. The possibilities are {0, 0, 6}; {0, 1, 5}; {0, 2, 4}; {0, 3, 3}; {1, 1, 4}; {1, 2, 3}; {2, 2, 2} i.e. 7 possibilities.

7 is our answer because each possible way of grouping can be done in only 1 way. Why? Because all hats are identical, so “which hat is in which group” does not matter.

Thus, the answer is 7 ways.

Summary

Distributions of things help us to distribute things in different boxes. With the help of knowledge of the distribution of things, we can solve problems related to probability and statistics. Understanding of Distribution of things helps us to solve problems related to real-life applications.

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Solved Example Based on Distribution Of Things

Example 1: The number of ways to distribute 30 identical candies among four children $\mathrm{C}_1, \mathrm{C}_2, \mathrm{C}_3$ and $\mathrm{C}_4$ so that $\mathrm{C}_2$ receives at least 4 and at most 7 candies, $\mathrm{C}_3$ receives at least 2 and at most 6 candies, is equal to:
[JEE MAINS 2022]
Solution: $4 \leq \mathrm{c}_2 \leq 7$

$
2 \leq \mathrm{c}_3 \leq 6
$

Now for the solution of $\mathrm{C}_1+\mathrm{C}_2+\mathrm{C}_3+\mathrm{C}_4=30$ with given conditions, the number of solutions equal
coefficient of $x^{30}$ in $\left(1+x+x^2+\cdots\right)\left(x^4+x^5+x^6+x^7\right)$ $\left(x^2+x^3+x^4+x^5+x^6\right)\left(1+x+x^2+\ldots\right)$
$=$ coefficient of $x^{30}$ in $x^6\left(1+x+x^2+\cdots\right)^2$
$\left(1+x+x^2+x^3\right)\left(1+x+x^2+\ldots+x^4\right)$
$=$ coofficient of $x^{24}$ in $\frac{1}{(1-x)^2} \cdot \frac{\left(x^4-1\right)}{(x-1)} \cdot \frac{\left(x^5-1\right)}{(x-1)}$
$=$ coefficient of $x^{24}$ in $\left(x^4-1\right)\left(x^5-1\right)(1-x)^{-4}$

$\begin{aligned} & =\text { coefficient of } \mathrm{x}^{24} \text { in }\left(\mathrm{x}^9-\mathrm{x}^5-\mathrm{x}^4+1\right)\left(1+4 \mathrm{x}+{ }^5 \mathrm{C}_2 \mathrm{x}^2+\ldots \ldots\right) \\ & ={ }^{18} \mathrm{C}_{15}-{ }^{22} \mathrm{C}_{19}-{ }^{23} \mathrm{C}_{20}+{ }^{27} \mathrm{C}_{24} \\ & ={ }^{18} \mathrm{C}_3-{ }^{22} \mathrm{C}_3-{ }^{23} \mathrm{C}_3+{ }^{27} \mathrm{C}_3 \\ & =816-1540-1771+2925 \\ & =430\end{aligned}$

Hence, the answer is 430

Example 2: The number of ways in which the examiner can assign 30 marks to 8 questions, giving not less than 2 marks to any questions, is : [JEE MAINS 2013]

Solution

Solution
${2} {2}{2}{2}{2}{2}{2}{2} \Rightarrow 30-16=14$ Mark

Hence,

$
\begin{aligned}
& n=8 \\
& r=14
\end{aligned}
$
$
\text { no of ways }={ }^{n+r-1} C_r={ }^{8+14-1} C_{14}={ }^{21} C_{14}
$
Hence, the answer is ${ }^{21} C_7$

Example 3: In how many ways can 5 different books can be tied up in 4 bundles?

Solution: The number of ways in which n different things can be distributed into r different groups is

$r^n-{ }^r C_1(r-1)^n+{ }^r C_2(r-2)^n-{ }^r C_3(r-3)^n+\ldots \ldots+(-1)^{r-1} \cdot{ }^r C_{r-1}$

or

$\sum_{p=0}^r(-1)^p \cdot r C_p \cdot(r-p)^n$

The number of ways $=\frac{1}{4!}\left(4^5-{ }^4 C_1 \cdot 3^5+{ }^4 C_2 \cdot 2^5-{ }^4 C_3 \cdot 1^5\right)=10$

Hence, the answer is 10

Example 4: The number of ways in which 6 different balls can be put in two boxes of different sizes so that no box remains empty is:

Solution: There are two distinct boxes for 6 different balls so each ball has 2 choices
\text { So we have } 2^6
But no box remains empty,
So number of ways $=2^6-2=62$

Hence, the answer is 62

Example 5: In how many ways can 7 different prizes be distributed among 3 students, where each student can receive at most one prize and one student gets two prizes?

Solution: To calculate the number of ways the 7 different prizes can be distributed among 3 students, where each student can receive at most one prize and one student gets two prizes, we can consider the following:

First, we select the students who will receive two prizes. There are 3 students to choose from, so there are 3 options.

Next, we assign the two prizes to the chosen student. There are 7 prizes to choose from for the first prize and 6 remaining prizes to choose from for the second prize. This can be calculated as $7 \times 6=42$ options.

Finally, we distributed the remaining 5 prizes among the other two students. Each of the remaining two students can receive at most one prize, so for each prize, there are 2 students to choose from.
\text { Since there are } 5 \text { remaining prizes, this gives us } 2^5=32 \text { options. }

To calculate the total number of ways to distribute the prizes, we multiply the number of options for each step:

$3 \times 42 \times=4032$

Hence, there are 4,032 distinct ways to distribute the 7 different prizes among 3 students, where each student can receive at most one prize and one student gets two prizes.

Hence, the answer is 4032