IPMAT Indore 2024 · Quantitative Ability · Question 35

Step 1 — shift each variable to a zero floor. Define

{y}_{1}={x}_{1}-1\ge 0,{y}_{2}={x}_{2}-2\ge 0,{y}_{3}={x}_{3}\ge 0,{y}_{4}={x}_{4}\ge 0

Step 2 — rewrite the equation.

\left({y}_{1}+1 \right)+\left({y}_{2}+2 \right)+{y}_{3}+{y}_{4}=50 \Rightarrow{y}_{1}+{y}_{2}+{y}_{3}+{y}_{4}=47

Step 3 — apply stars and bars. The number of non-negative integer solutions of {y}_{1}+\cdots +{y}_{k}=N is \left( \genfrac{}{}{0pt}{}{N+k-1}{k-1}\right). With N=47, k=4:

\left( \genfrac{}{}{0pt}{}{47+3}{3}\right)=\left( \genfrac{}{}{0pt}{}{50}{3}\right)

Step 4 — evaluate.

\left( \genfrac{}{}{0pt}{}{50}{3}\right)=\frac{50\times 49\times 48}{6}=\frac{117600}{6}=19600

Answer: Option 1 — 19600

Shortcut. Subtract the floors first, then one formula. The lower bounds cost 1+2+0+0=3, so the problem becomes {y}_{1}+{y}_{2}+{y}_{3}+{y}_{4}=47 with all {y}_{i}\ge 0, giving \left( \genfrac{}{}{0pt}{}{50}{3}\right).

Fast evaluation: \frac{50\cdot 49\cdot 48}{6} — cancel 48/6=8 first, then 50\times 49\times 8=19600. Always cancel before multiplying; it keeps every intermediate under three digits.

Note that \left( \genfrac{}{}{0pt}{}{50}{3}\right) is a number worth recognising, as it recurs across IPMAT and CAT papers.