JEE Advanced · 2009 · 12 Apr 2009 · Paper 1 · Matrices

A $3\times 3$ symmetric matrix has 6 independent entries (3 diagonal + 3 above diagonal). We need to choose 5 positions for 1's from these 6 positions.

However, we need to count more carefully:

• If 2 zeros are on the diagonal: $\left(\genfrac{}{}{0ex}{}{3}{2}\right)\times \left(\genfrac{}{}{0ex}{}{3}{1}\right)=3\times 3=9$ ways (choose 2 diagonal positions for 0, then 1 off-diagonal position for the remaining 0)
• If 1 zero is on the diagonal: $\left(\genfrac{}{}{0ex}{}{3}{1}\right)\times \left(\genfrac{}{}{0ex}{}{3}{2}\right)=3\times 3=9$ ways... but this gives 5 ones on diagonal + off-diagonal, which doesn't work.
• If 0 zeros on diagonal (all diagonal entries are 1): $\left(\genfrac{}{}{0ex}{}{3}{1}\right)=3$ ways (choose 1 off-diagonal position for 0)

Actually: diagonal has 3 entries, off-diagonal has 3 independent entries. Total 6 independent entries. Choose 4 for zeros: but we need exactly 5 ones.

Correct count: $\left(\genfrac{}{}{0ex}{}{3}{2}\right)\times 3+\left(\genfrac{}{}{0ex}{}{3}{0}\right)\times \left(\genfrac{}{}{0ex}{}{3}{1}\right)=9+3=12$