JEE Advanced 2023 · Paper 2, Mathematics question 10

Multiples of 5. No 5 is available, so a multiple of 5 must end in 0. The first four digits then come from the multiset $\{1,2,2,2,4,4\}$, and there is no leading-zero problem.

Count words of length 4 with at most one 1, three 2s and two 4s:

(#1, #2, #4)	(0,2,2)	(0,3,1)	(1,1,2)	(1,2,1)	(1,3,0)
count	6	4	12	12	4

Total: 38.

Multiples of 20 also need an even tens digit.

• Tens digit 2: the first three digits come from $\{1,2,2,4,4\}$: $3+3+3+3+6=18$ ways.
• Tens digit 4: the first three digits come from $\{1,2,2,2,4\}$: $1+3+3+6=13$ ways.

Total: 31.

$$p=\frac{31}{38},\qquad38p=31.$$