IPMAT Indore 2023 · Quantitative Ability · Question 12

Step 1 — recall the count for a fixed sum. The number of solutions of a+b+c=s in positive integers is

\left( \genfrac{}{}{0pt}{}{s-1}{2}\right)

(Place s identical items in a row and choose 2 of the s-1 gaps between them.)

Step 2 — sum over the allowed totals. The condition 21\le a+b+c\le 25 allows s=21,22,23,24,25:

\left( \genfrac{}{}{0pt}{}{20}{2}\right)+\left( \genfrac{}{}{0pt}{}{21}{2}\right)+\left( \genfrac{}{}{0pt}{}{22}{2}\right)+\left( \genfrac{}{}{0pt}{}{23}{2}\right)+\left( \genfrac{}{}{0pt}{}{24}{2}\right)

Step 3 — evaluate each.

190+210+231+253+276

Step 4 — add.

190+210=400; 400+231=631; 631+253=884; 884+276=1160

Answer: 1160

Shortcut. Use the hockey-stick identity instead of adding five terms.

\sum_{s=21}^{25}{\left( \genfrac{}{}{0pt}{}{s-1}{2}\right)}=\sum_{m=20}^{24}{\left( \genfrac{}{}{0pt}{}{m}{2}\right)}=\left( \genfrac{}{}{0pt}{}{25}{3}\right)-\left( \genfrac{}{}{0pt}{}{20}{3}\right)

=2300-1140=1160

Two binomial evaluations replace five, and \left( \genfrac{}{}{0pt}{}{25}{3}\right) and \left( \genfrac{}{}{0pt}{}{20}{3}\right) are both quick mental computations.

Watch the inclusivity. If the bounds were strict (21<a+b+c<25) only s=22,23,24 would count, giving 694. The difference between 694 and 1160 is entirely about whether the endpoints are included — read the inequality signs before you start.