Question

The number of ordered triplets of positive integers which are solutions of the equation $x + y + z = 100$ is?
$1)6005$
$2)4851$
$3)5081$
$4)$None of these

Answer 1

Hint: First, to solve the given question we will make use of the partition method which says that the number of ways of dividing the $n$identical objects into $r$distinct groups where each group can get any number of objects.
Here the groups are $3$because we have three unknown variables and the things we will need to distribute are $100$(we need to put a hundred things into the three groups or boxes).

Formula used:
The General combination formula is ${}^n{c_r} = \dfrac{{n!}}{{r!(n - r)!}}$
Each group gets at least one object in the combination formula is ${}^{n + r - 1}{C_{r - 1}}$, where r is the distinct groups and n is the objects that are given.


Complete step by step answer:
The question asks us the find the values of $(x,y,z)$such that their sum is $100$
We can say that we have to divide $100$into three groups $(x,y,z)$each group at least one object.
From the given we have, $x + y + z = 100$
We cannot use the partition method because $(x,y,z)$are given as the natural numbers, the groups can also get $0$things.
Thus, we will do some changes in the equation as $x = a + 1,y = b + 1,z = c + 1$(positive integers)
Then we get the original equation as $a + b + c + 3 = 100 \Rightarrow a + b + c = 97$
Here, $(a,b,c)$are whole numbers. thus, we can apply the partition method now.
Now we have $97$things into $3$groups.
Hence the total ways are, ${}^{n + r - 1}{C_{r - 1}} = {}^{97 + 3 - 1}{C_{3 - 1}} \Rightarrow {}^{99}{C_2}$
From the combination general formula we get,${}^{99}{C_2} = \dfrac{{99!}}{{2!(99 - 2)!}}$ (by the use of combination formula)
Further solving this we get, ${}^{99}{C_2} = \dfrac{{99!}}{{2!(99 - 2)!}} \Rightarrow \dfrac{{99!}}{{2!97!}} \Rightarrow \dfrac{{99 \times 98 \times 97!}}{{2!97!}}$
\[{}^{99}{C_2} = \dfrac{{99 \times 98 \times 97!}}{{2!97!}} \Rightarrow \dfrac{{99 \times 98}}{{2!}}\](Cancelling the common terms)
 ${}^{99}{C_2} = \dfrac{{99 \times 98}}{2}$
Thus, we bet, ${}^{99}{C_2} = \dfrac{{99 \times 98}}{2} \Rightarrow 4851$is the required number of ordered triplets of positive integers.
Hence option $2)4851$is correct.

Note: We can also able to solve the problem without converting it into the whole number as ${}^{n + r - 1}{C_{r - 1}} = {}^{100 + 3 - 1}{C_{3 - 1}} \Rightarrow {}^{102}{C_2} \Rightarrow 5151$
Now apply the variables to zero one by one we get;
For $x = 0$then we get, the total number of ways is $100$
For $y = 0$ then we get, the total number of ways is $100$
For $z = 0$ then we get, the total number of ways is $100$
Required number of ways = total numbers – variables zero terms.
Thus, we get, the required number of ways = $5151 - (100 + 100 + 100)$
Hence, we get the required number of ways $ = 4851$
Factorial can be expressed in the form of $n! = n \times (n - 1)! \times ... \times 2 \times 1$