If all the digits in the number 53426 are per | vedclass

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Question

(A) The digits are $\{2, 3, 4, 5, 6\}$. Arranging them in decreasing order: $6, 5, 4, 3, 2$. Total permutations possible = $5! = 120$. We want the $89

Vedclass · Accepted Answer

$34265$

Vedclass · Answer

(A) The digits are $\{2, 3, 4, 5, 6\}$. Arranging them in decreasing order: $6, 5, 4, 3, 2$. Total permutations possible = $5! = 120$. We want the $89^{th}$ number when arranged in decreasing order. Numbers starting with $6$: $4! = 24$ numbers (Ranks $1$ to $24$). Numbers starting with $5$: $4! = 24$ numbers (Ranks $25$ to $48$). Numbers starting with $4$: $4! = 24$ numbers (Ranks $49$ to $72$). Numbers starting with $36$: $3! = 6$ numbers (Ranks $73$ to $78$). Numbers starting with $35$: $3! = 6$ numbers (Ranks $79$ to $84$). Numbers starting with $346$: $2! = 2$ numbers (Ranks $85$ to $86$). Numbers starting with $345$: $2! = 2$ numbers (Ranks $87$ to $88$). The $89^{th}$ number starts with $34265$ (Rank $89$) and the $90^{th}$ is $34256$ (Rank $90$). Thus,the $89^{th}$ number is $34265$.

If all the digits in the number $53426$ are permuted in all possible ways and are arranged in decreasing order,then the number having rank $89$ is:

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