Question:
The number of ordered triplets of positive integers which are solution of the equation x y z=100 is (solve using binomial theorem)
Answer:
Given, x + y + z = 100;
where 0 < x < 99, 0 < y < 99, 0 < z < 99
x , y, and z can only be from 1 to 98
Because if the other two numbers are both 1, the value of the third is 98.
if x = 1, y could only be 1 to 98 (98 choices)
if x = 2, y could only be 1 to 97 (97 choices)
......
If x = 98, c could only be 1 (1 choice)
Once x and y are selected, z could only be 100 - (x+y)
Se we only count the various x, y combinations
Total number of ordered triplets
= 98 + 97 + .... + 2 + 1
= 98(98 + 1)/2 = 4851
OR
Given, x + y + z = 100;
where x ≥ 1, y ≥ 1, z ≥ 1
Let u = x - 1, v = y - 1, w = z - 1
where u ≥ 0, v ≥ 0, w ≥ 0
Now, equation becomes
u + v + w = 97
So, the total number of solution = 97+3-1C3-1
= 99C2
= (99 * 98)/2
= 4851
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