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Question: The number of solutions of the equation x + y + z = 10 in positive integers x, y, z is equal to: (...

The number of solutions of the equation x + y + z = 10 in positive integers x, y, z is equal to:
(a) 36
(b) 55
(c) 72
(d) 45

Explanation

Solution

Hint: To solve the question given above, we will make use of partition method which says that the number of ways of dividing ‘n’ identical objects into ‘r’ distinct groups where each group can get any number of objects is given by n+r−1Cr−1.^{n+r-1}C{_{r-1}}. Here, there are 3 groups in which we want to distribute 10 things.

Complete step by step solution:
The question says that we have to find the number of values of (x, y, z) such that their sum is 10. In other words, we can say that we have to divide 10 into three groups x, y, and z such that each group gets at least one object. So, we will solve this question with the help of the partition method. The partition method says that the number of ways of dividing ‘n’ identical objects into ‘r’ distinct groups where each group can get any number of objects is given by n+r−1Cr−1.^{n+r-1}C{_{r-1}}. In our case, we have,
x+y+z=10....(i)x+y+z=10....\left( i \right)
We cannot use the partition method directly here because x, y, and z are natural numbers in our case. But for us to apply the partition method, the groups can also get 0 things. Thus, we will do some changes in equation (i). We will assume x = a + 1, y = b + 1 and z = c + 1. So, the equation (i) becomes,
⇒(a+1)+(b+1)+(c+1)=10\Rightarrow \left( a+1 \right)+\left( b+1 \right)+\left( c+1 \right)=10
⇒a+b+c+3=10\Rightarrow a+b+c+3=10
⇒a+b+c=7\Rightarrow a+b+c=7
Here, a, b and c are whole numbers. Thus, we can apply the partition method now. Here, we have to divide 7 things into 3 groups. The number of ways is given by,
Total ways= 7+3−1C3−1= 9C2\text{Total ways}={{\text{ }}^{7+3-1}}{{C}_{3-1}}={{\text{ }}^{9}}{{C}_{2}}
⇒Total ways=9!(9−2)!2!\Rightarrow \text{Total ways}=\dfrac{9!}{\left( 9-2 \right)!2!}
⇒Total ways=9!7!2!\Rightarrow \text{Total ways}=\dfrac{9!}{7!2!}
⇒Total ways=9×8×7!7!2!\Rightarrow \text{Total ways}=\dfrac{9\times 8\times 7!}{7!2!}
⇒Total ways=9×82!\Rightarrow \text{Total ways}=\dfrac{9\times 8}{2!}
⇒Total ways=36\Rightarrow \text{Total ways}=36
Hence, option (a) is the right answer.

Note: The alternate method to solve the above question is by first finding the total number of ways of dividing the objects and then subtracting those cases from it when x or y or z is 0. The total number of ways of obtaining x + y + z = 0 is 10+3−1C3−1=66.^{10+3-1}{{C}_{3-1}}=66. Now there are three undesired cases in it:
Case 1: When x = 0, there will be a total of 10 ways.
Case 2: When y = 0, there will be a total of 10 ways.
Case 3: When z = 0, there will be a total of 10 ways.
Thus, we have,
Required Number of Ways = Total Ways – (Case 1 + Case 2 + Case 3)
Required Number of Ways = 66 – (10 + 10 + 10)
Required Number of Ways = 36