Question

How many different ways can $15$ candy bars be distributed between Ram, Shyam, Ghanshyam and Balram, if Ram cannot have more than $5$ candy bars and Shyam must have at least two? Assume all candy bars to be alike.

Answer 1

Hint: This question can be solved using the concept of combinations.
Combinations in mathematics can be simply explained as a mathematical technique that determines the number of arrangements possible for collection of items where the order of the selection does not matter. So in combinations we can select an item in any order.
So by using this concept we can solve the above given question.

Complete step by step answer:
Given
$
  {\text{Ram}} + {\text{Shyam}} + {\text{Ghanshyam}} + {\text{Balram}} = {\text{15 candy bars}}.............................\left( i \right) \\
  {\text{Ram}} \leqslant {\text{5}}....................................\left( {ii} \right) \\
  {\text{Shyam}} \geqslant {\text{2}}...............................\left( {iii} \right) \\
 $
Now we have to find how these $15$ candies can be distributed.
Now we know that Shyam must have at least $2$ candies.
So let’s first give Shyam $2$ candies.
So now the new total number of candies would be:
${\text{Ram}} + {\left( {{\text{Shyam}}} \right)^\prime } + {\text{Ghanshyam}} + {\text{Balram}} = {\text{13 candy bars}}$
Now let’s take the 2nd condition: Ram cannot have more than $5$ candies.
So let’s take various cases and thus find the corresponding combinations:
1st case: Ram has 0 candies:
$ \Rightarrow {\left( {{\text{Shyam}}} \right)^\prime } + {\text{Ghanshyam}} + {\text{Balram}} = {\text{13 candy bars}}$
So the number of ways the candies can be distributed is given by:
$^{13 + 3 - 1}{C_{3 - 1}}{ = ^{15}}{C_2} = \dfrac{{15!}}{{2!\left( {15 - 2} \right)!}} = 105$
2nd case: Ram has 1 candy:
$ \Rightarrow {\left( {{\text{Shyam}}} \right)^\prime } + {\text{Ghanshyam}} + {\text{Balram}} = {\text{12 candy bars}}$
So the number of ways the candies can be distributed is given by:
\[^{12 + 3 - 1}{C_{3 - 1}}{ = ^{14}}{C_2} = \dfrac{{14!}}{{2!\left( {14 - 2} \right)!}} = 91\]
3rd case: Ram has 2 candies:
$ \Rightarrow {\left( {{\text{Shyam}}} \right)^\prime } + {\text{Ghanshyam}} + {\text{Balram}} = {\text{11 candy bars}}$
So the number of ways the candies can be distributed is given by:
$^{11 + 3 - 1}{C_{3 - 1}}{ = ^{13}}{C_2} = \dfrac{{13!}}{{2!\left( {13 - 2} \right)!}} = 78$
4th case: Ram has 3 candies:
$ \Rightarrow {\left( {{\text{Shyam}}} \right)^\prime } + {\text{Ghanshyam}} + {\text{Balram}} = {\text{10 candy bars}}$
So the number of ways the candies can be distributed is given by:
$^{10 + 3 - 1}{C_{3 - 1}}{ = ^{12}}{C_2} = \dfrac{{12!}}{{2!\left( {12 - 2} \right)!}} = 66$
5th case: Ram has 4 candies:
$ \Rightarrow {\left( {{\text{Shyam}}} \right)^\prime } + {\text{Ghanshyam}} + {\text{Balram}} = 9{\text{ candy bars}}$
So the number of ways the candies can be distributed is given by:
$^{9 + 3 - 1}{C_{3 - 1}}{ = ^{11}}{C_2} = \dfrac{{11!}}{{2!\left( {11 - 2} \right)!}} = 55$
6th case: Ram has 5 candies:
\[ \Rightarrow {\left( {{\text{Shyam}}} \right)^\prime } + {\text{Ghanshyam}} + {\text{Balram}} = 8{\text{ candy bars}}\]
So the number of ways the candies can be distributed is given by:
$^{8 + 3 - 1}{C_{3 - 1}}{ = ^{10}}{C_2} = \dfrac{{10!}}{{2!\left( {10 - 2} \right)!}} = 45$
Now the total number of ways of distribution would be the sum total of all the combinations of each case.
Which would be:
$105 + 91 + 78 + 66 + 55 + 45 = 440$.
Therefore there can be $440$ ways in which $15$ candy bars distributed between Ram, Shyam, Ghanshyam and Balram, where Ram cannot have more than $5$ candy bars and Shyam must have at least two.

Note:
General formula to find combination in the form $^n{C_r}$is:
 $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
While doing a question involving combinations one should be careful about the calculation processes since it can be tricky as well as lengthy.
Also the above given method is recommended to solve similar questions involving combinations.