GATE 2023 IN Q.49: solution and answer

Official answer: 240Source: Official answer key, GATE 2023 IN (IIT Kanpur)

1. What the number looks like. Read the condition the standard way: one digit is repeated (it appears twice) and no other digit repeats. A five-digit number has 5 places. The repeated digit fills 2 of them, and the other 3 places hold three different digits, none equal to the repeated one. Only 3 other digits are left in $\{3,4,5,6\}$, so all three are used once each ($2+1+1+1=5$). Every such number therefore uses all four digits, and exactly one of them appears twice.

2. Choose the repeated digit. Any of the 4 digits can be the one that appears twice: $\left(\genfrac{}{}{0ex}{}{4}{1}\right)=4$ ways.

3. Arrange the multiset. Once the repeated digit is fixed, the five digits form a multiset with one item appearing twice. Its number of distinct orders is $\frac{5!}{2!}=\frac{120}{2}=60$ (divide by $2!$ because swapping the two equal digits gives the same number).

4. Leading digit. None of the digits is $0$, so every order is a genuine five-digit number and nothing has to be subtracted.

5. Multiplication principle. $$N=4\times \frac{5!}{2!}=4\times 60=240$$

6. Check by placing the digits. Choose the repeated digit (4 ways). Choose its two places: $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$. Put the other three digits in the remaining 3 places: $3!=6$. That gives $4\times 10\times 6=240$, the same count.