This is a question from CAT permutation and combination. This question is more like a puzzle than a question. But an interesting puzzle nevertheless. We have to find the positive integral solutions of a set of variables. But there's a constraint that governs these variables. CAT Permutation and Combination and Probability is an important topic in the**CAT Exam**. One can expect to get 2-3 questions from this topic in ** CAT exam**.

Question 25 : The number of solutions (x, y, z) to the equation x – y – z = 25, where x, y, and z are positive integers such that x ≤ 40, y ≤ 12, and z ≤ 12 is

- 101
- 99
- 87
- 105

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We have to find the number of solutions (x, y, z) to the given equation.

x – y – z = 25, where x, y, and z are positive integers such that x ≤ 40, y ≤ 12, and z ≤ 12.

x – y – z = 25, so the maximum value y and z can take is 12 and x can take is 40.

From 40 we can subtract some values to get 25 i.e.15 can be subtracted,

If x = 40 then y + z = 15

y ≤ 12 so y can take all the corresponding values 12,11,10,......3 it is a set of 10 values.

z ≤ 12, so z can take corresponding other values to get with 15 when added with x

Similarly,

If x = 39, then y + z = 14 here y can take all the value from 12 ,11,...till 2

So there are 11 values.

If x = 38, then y + z = 13 here y can take all values from 12,11,10,....till 1

So there are 12 values.

If x = 37 then y + z = 12 here y can take 11,10,9,.....,1 so there are 11 values

If x = 36 then y + z = 11 here y can take 10,9,......,1 so there are 10 values

If x = 35 then y + z = 10 here y can take 9,8,.......,1 so there are 9 values

If x = 34 then y + z = 9 here y can take 8,.......,1 so there are 8 values

If x = 33 then y + z = 8 here y can take 7,........,1 so there are 7values

If x = 32 then y + z = 7 here y can take 6,........,1 so there are 6 values

If x = 31 then y + z = 6 here y can take 5.........,1 so there are 5 values

If x = 30 then y + z = 5 here y can take 4,......,1 so there are 4 values

If x = 29 then y + z = 4 here y can take 3,...,1 so there are 3values

If x = 28 then y + z = 3 here y can take 2,1 so there are 2 values

If x = 27 then y + z = 2 here y can take 1 so there is 1 value

x cannot be less than 27 because it is given that y and z are positive integers so it has to be at least one cannot be less than one

So, \\frac{12 × 13}{2}) + 21 = 78 + 21 = 99.

The number of solutions (x, y, z) to the equation x – y – z = 25, where x, y, and z are positive integers such that x ≤ 40, y ≤ 12, and z ≤ 12 is 99.

The question is **"The number of solutions (x, y, z) to the equation x – y – z = 25, where x, y, and z are positive integers such that x ≤ 40, y ≤ 12, and z ≤ 12 is" **

Choice B is the correct answer

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