Distribution Of Things - Practice Questions & MCQ

No restriction

To distribute 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....... times $=r^n$

Non-empty restriction

To distribute 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 into $r$ 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 objects can be given as these are identical), then we can distribute the remaining ( n -r) objects in r people in ${ }^{n-r+r-1} C_{r-1}={ }^{n-1} C_{r-1}$ ways.

Alternative:

This thing can also be comprehended as the following

Let the first group gets $\mathrm{a}_1$ objects
Second group gets $\mathrm{a}_2$ objects
$\qquad$
$r^{\text {th }}$ group gets $\mathrm{a}_{\mathrm{r}}$ objects
So, $a_1+a_2+a_3+\ldots+a_r=n$ $\qquad$
Now if empty groups are allowed we need to find the whole number solution of the equation (i)
So, whole number solutions of $a_1+a_2+a_3+\ldots+a_r=n$ equals 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 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: $1^{\text {st }}=113,2^{\text {nd }}=122$

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 $\mathbf{2}^{\text {nd }}$ 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.