IPMAT Indore 2022 · Quantitative Ability · Question 14

Step 1 — count arrangements with the Sharmas together. Glue the Sharmas into a single block. That leaves 8-2+1=7 entities around a round table:

\left( 7-1 \right)!=6!=720

and the Sharmas can sit within their block in 2!=2 orders:

720\times 2=1440

Step 2 — count arrangements with both couples together. Glue both couples, leaving 8-4+2=6 entities:

\left( 6-1 \right)!=5!=120

each couple internally arrangeable in 2 ways:

120\times 2\times 2=480

Step 3 — subtract. “Sharmas together and Ahujas apart” is “Sharmas together” minus “both together”:

1440-480=960

Answer: 960

Shortcut. Glue, then subtract — never try to place the Ahujas apart directly.

\underset{\text{Sharmas together}}{\underset{⏟}{6!\times 2}}- \underset{\text{both couples together}}{\underset{⏟}{5!\times 2\times 2}}= 1440-480=960

The circular-table rule is where marks are lost: n entities around a round table give \left( n-1 \right)! arrangements, not n!, because rotations are identical. Count your entities after gluing — 7 in the first case, 6 in the second.

A quick sanity check: the answer must be less than 1440 and a clear majority of it, since gluing a second specific couple is restrictive. 960=2/3\times 1440 sits exactly where you would expect.