GATE 2026 DA Q.20: solution and answer

GATE 2026 · DAQ.20 · 1 mark · MCQ

Suppose that a computer program provides a non-negative and integer-valued random solution to the equation $n_{1} + n_{2} + n_{3} + n_{4} = 20$ . Which of the following is the probability that all of $n_{1}, n_{2}, n_{3}, n_{4}$ in the provided solution are positive?

(A)
$(3 19) / (3 23)$
(B)
$(4 20) / (4 24)$
(C)
$(3 20) / (3 23)$
(D)
$(4 19) / (4 24)$

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Answer and solution

ANSWER

The probability is $(3 19) / (3 23)$ , option (A).

Official answer: ASource: Official answer key, GATE 2026 DA (IIT Guwahati)

The question treats the program's output as a random solution chosen uniformly from all non-negative integer solutions of $n_{1} + n_{2} + n_{3} + n_{4} = 20$ . This is the usual reading, so the probability is $\frac{favourable solutions}{total solutions}$ .
Total: by stars and bars, the number of non-negative integer solutions of $n_{1} + n_{2} + n_{3} + n_{4} = 20$ is $(4 - 1 20 + 4 - 1) = (3 23) = 1771$ .
Favourable: put $m_{i} = n_{i} - 1$ for each $i$ . Then $n_{i} \geq 1$ exactly when $m_{i} \geq 0$ , and the equation becomes $m_{1} + m_{2} + m_{3} + m_{4} = 20 - 4 = 16$ . Its number of non-negative solutions is $(3 16 + 3) = (3 19) = 969$ .
Probability $= \frac{( 3 19 )}{( 3 23 )} = \frac{969}{1771} \approx 0.547$ , which is option (A).
Check on the other options. (B) and (D) have denominator $(4 24)$ . That would count non-negative solutions of a sum with five variables, or of $n_{1} + \dots + n_{4} = 20$ with $n_{i} \leq$ something, and neither applies here. (C) has numerator $(3 20)$ , which is the number of solutions with all $n_{i} \geq 1$ of a sum equal to 21, not 20.

KEY IDEA

Count with stars and bars. All non-negative solutions: $(k - 1 n + k - 1)$ . For positive solutions, subtract 1 from each variable first, so the sum falls from $n$ to $n - k$ .

WATCH OUT FOR

Using $(3 20)$ or $(4 19)$ for the count, which comes from mixing up " $n + k - 1$ choose $k - 1$ " with " $n$ choose $k$ ". Another slip is forgetting to subtract 4 from 20 for the positive case.

From the GATE 2026 DA paper. Every question of this paper · Download the full paper · All 5 Counting & axioms questions, with notes

GATE 2026 DA Q.20: solution and answer

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