JEE Advanced 2023 Paper 2 · Q10 · Permutations & Combinations

Question

Let $X$ be the set of all five digit numbers formed using digits $1, 2, 2, 2, 4, 4, 0$. (For example $22240$ is in $X$ while $02244$ and $44422$ are not in $X$.) Suppose that each element of $X$ has an equal chance of being chosen. Let $p$ be the conditional probability that an element chosen at random is a multiple of $20$ given that it is a multiple of $5$. Then the value of $38p$ is equal to ___.

SolveKar AI · Accepted Answer

An element of $X$ is a five-digit number with exactly the digits drawn (no leading zero). Multiples of $5$ end in $0$ (the digit $5$ isn't in the multiset). Count multiples of $5$ (last digit $0$): the remaining $4$ digits chosen from $\{1,2,2,2,4,4\}$, valid 4-digit selections: $\{1,2,2,2\}$ → $4!/3! = 4$; $\{1,4,2,2\}$ → $4!/2! = 12$; $\{1,2,4,4\}$ → $4!/2! = 12$; $\{4,2,2,2\}$ → $4!/3! = 4$; $\{2,2,4,4\}$ → $4!/(2!2!) = 6$. Total = $4+12+12+4+6 = 38$. Multiples of $20$ end in $00$ or $20$ or $40$ or $60$ or $80$. Since last digit is $0$, the second-last must be even (i.e. $2$ or $4$). Subtract cases where second-last is $1$: those are $\{2,2,2,1\}$ ending in $\ldots 10$: 1 perm; $\{4,2,2,1\}$ ending in $\ldots 10$: $3!/2! = 3$; $\{2,4,4,1\}$ ending in $\ldots 10$: $3!/2! = 3$. So multiples of $20$ count = $38 - (1+3+3) = 31$. So $p = 31/38$ and $38p = 31$.

JEE Advanced 2023 Paper 2 · Q10 · Permutations & Combinations

Solution