IPMAT Indore 2023 · Quantitative Ability · Question 15

Step 1 — name the coefficients. Let them be \left( \genfrac{}{}{0pt}{}{n}{r-1}\right), \left( \genfrac{}{}{0pt}{}{n}{r}\right), \left( \genfrac{}{}{0pt}{}{n}{r+1}\right) with

\left( \genfrac{}{}{0pt}{}{n}{r-1}\right):\left( \genfrac{}{}{0pt}{}{n}{r}\right):\left( \genfrac{}{}{0pt}{}{n}{r+1}\right)=1:9:63

Step 2 — use the ratio of consecutive binomial coefficients. The standard identity is

\frac{\left( \genfrac{}{}{0pt}{}{n}{r}\right)}{\left( \genfrac{}{}{0pt}{}{n}{r-1}\right)}=\frac{n-r+1}{r}

From the first pair \left( 9/1=9 \right):

\frac{n-r+1}{r}=9 \Rightarrow n+1=10r

From the second pair \left( 63/9=7 \right):

\frac{n-r}{r+1}=7 \Rightarrow n=8r+7

Step 3 — solve the pair. From (1), n=10r-1. Substituting into (2):

10r-1=8r+7 \Rightarrow 2r=8 \Rightarrow r=4

n=10\left( 4 \right)-1=39

Step 4 — verify. With n=39, r=4:

\left( \genfrac{}{}{0pt}{}{39}{3}\right)=9139, \left( \genfrac{}{}{0pt}{}{39}{4}\right)=82251, \left( \genfrac{}{}{0pt}{}{39}{5}\right)=575757

82251=9\times 9139 ✓ 575757=7\times 82251 ✓

Answer: 39

Shortcut. Never compute a binomial coefficient — only their ratios.

\frac{\left( \genfrac{}{}{0pt}{}{n}{r}\right)}{\left( \genfrac{}{}{0pt}{}{n}{r-1}\right)}=\frac{n-r+1}{r}

Apply it twice, to 9:1 and to 63:9=7:1, giving two linear equations in n and r. Subtracting them gives r=4 and n=39 in about four lines.

Note the structure the setter used: the ratio 1:9:63 is really 1:9:9\cdot 7, so the two successive ratios are 9 and 7. Always convert a three-term ratio into two consecutive ratios first — that is the form the identity consumes.

(The supplied file tags this as Data Interpretation; it is Binomial Theorem, and is retagged here.)