Problems of kind “In how many ways we can distribute n identical coins among r Beggars?” in Permutation and combination is called beggars problem.

To solve these kind of problem there is a simple logic:

given in the video :

Other variations to this problem are:

##VARIATION 1::Find the number of non negative integer solutions of the equation X + Y + Z = 8 ?

or Find the number of distinct terms in the expansion of (x+ y+ z)^8 ?

In this problem X,Y,Z are consider as beggars and ‘8’ as 8 identical coins.

Solution: This question is same as the question we solved above

Note that all X , Y, Z ≥ 0 and all are integers.

Here 8 on the right hand side represents the number of coins and X , Y and Z represents 3 beggars . Lets us take the same cases which we took in previous example all these can be solutions of the above equations

1+ 3+ 4= 8

0 +2 +6 = 8

0+ 0+ 8 = 8

And many more such cases are possible. So clearly the number of non negative integer solutions of the equation X + Y + Z = 8 is same as the number of ways 8 identical one rupee coins can be distributed among 3 beggars =

Other similar problems:

or In how many ways 8 identical balls can be distributed into 3 different boxes so that no box remains empty?

or Three boys picked up 8 oranges. In how many ways can they divide them if all oranges be identical?

##VARIATION 2::Find the number of positive integer solutions of the equation X + Y + Z = 8?

This question is almost same but constraints are different i.e. X , Y, Z > 0

Or we can write X≥ 1, Y≥ 1, Z ≥ 1

Again the question is analogous to distributing 8 identical coins to three beggars

But here the condition is every beggar should have atleast 1 coin so we will distribute 1 coin to each beggar in advance.

So giving 1 coin to each beggar means we donated 3 coins. Thus we are left with 5 coins which we have to distribute among 3 beggars which can be done in

##VARIATION 3::Find the number of integral solutions of the equation X + Y + Z +W = 20 where X≥ 1 ,Y≥ 2, Z >0 , W ≥ 3?

Solution: This question is almost same but constraints are different i.e. X≥ 1 , Y≥ 2, Z >0 ,W ≥ 3

Again the question is analogous to distributing 20 identical coins to four beggars But here the condition is 1st beggar should have at least 1 coin, 2nd beggar should have at least 2 coin , 3rd beggar should have more than zero coin which is same as saying that he should have at least 1 coin, and 4th beggar should have at least 3 coins So we will give 1 coin to 1st beggar ,2 coins to 2nd beggar ,1 coin to 3rd beggar and 3 coins to 4th beggar first.

Now we are left with 13 coins to distribute among 4 beggars:

which is equals to

##VARIATION 4::How many non negative integral solutions are there to the system of equations X + Y + Z +W + K = 20 and W+ K = 7 ?

Solution: Putting the value of W+ K = 7 in

X + Y + Z +W + K = 20 when each of X≥ 0 , Y≥ 0 ,Z ≥ 0 ,W≥ 0 ,K≥ 0

We get X + Y + Z = 13

Solving this equation is same as distributing 13 Identical coins among 3 beggars which can be done in

Similarly solving equation W+ K = 7 where W≥ 0 ,K≥ 0 equivalent to distributing 7 Identical coins among 2 beggars which can be done in

Thus the total number of solutions to the system of equations is = 8 x 105=840

##VARIATION 5::QUESTION 8:Find the number of non negative integral solutions of 2X + Y + Z = 18 ?

Solution

The questions which we did till now had coefficient one but this has coefficient two. For the time being we put x =k so equation becomes

2K + Y +Z = 18

Y + Z = 18 – 2k (1)

As per the given constraint we are given Y ≥ 0 and Z ≥ 0 .

Thus Y + Z ≥ 0

=> 18 – 2k≥ 0

=> k≤ 9

Now as far as solving equation (1) is concerned we very well know it is equivalent to distributing 18 – 2k coins among 2 beggars

= 18- 2k + 1 = 19 – 2k ways

But k varies from 0 to 9.Thus total number of ways

##VARIATION 6::Find the number of non negative integral solutions of the equation X + Y + Z ≤18 ?

Solution

Given equation can be written as

X + Y + Z + W = 18 where each of X, Y, Z and W is greater than or equal to zero

Thus reduces to distributing 18 coins to 4 beggars which can be done

Hence the number of non-negative integral solutions of the equation X + Y + Z ≤ 18 are 1330.