The digits 1, 2, 3, 4, 5 are each used at most once to form a 2-digit number, so the two digits used must be different from each other. How many of these 2-digit numbers are odd?

AS Level Maths 9709: nPr with a Parity (Odd Number) Restriction

Step 1: Decide which position to fill first

For a number to be odd, its units (last) digit must be odd. Among $1,2,3,4,5$ , the odd digits are $1,3,5$ , that’s $3$ of them. It’s easiest to fill the units digit first, since it has the restriction, and then fill the tens digit from whatever remains.

Step 2: Count using the multiplication principle

Units digit: must be odd, so there are $3$ choices ( $1$ , $3$ or $5$ ).
Tens digit: must be different from the units digit (digits can’t repeat), so out of the $5$ digits, $1$ has already been used, leaving $4$ choices, and this count of $4$ is the same no matter which of the $3$ odd digits was chosen for the units place.

$3\times4 = 12.$

Step 3: Check by listing

Units $=1$ : tens can be $2,3,4,5$ , numbers $21,31,41,51$ .
Units $=3$ : tens can be $1,2,4,5$ , numbers $13,23,43,53$ .
Units $=5$ : tens can be $1,2,3,4$ , numbers $15,25,35,45$ .

That’s $4+4+4=12$ numbers in total, matching the count above.

Why the other options are wrong

A ( $9$ ): this comes from wrongly assuming only $3$ digits remain for the tens place (as if one of the $4$ leftover digits couldn’t be used), giving $3\times3=9$ instead of $3\times4=12$ .
C ( $15$ ): this comes from multiplying the $3$ odd digits by the full set of $5$ digits ( $3\times5=15$ ), rather than by the $4$ digits that actually remain once the units digit has been chosen.
D ( $20$ ): this is the total number of $2$ -digit numbers with distinct digits from $1$ – $5$ with no restriction on parity ( $5\times4=20$ ). The “odd” condition was never applied.

Final answer

Number of $2$ -digit numbers that are odd $= \boxed{12}$ , option B.