The number of four-digit numbers strictly gre | vedclass

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(C) We need to form a $4$-digit number $d_1 d_2 d_3 d_4$ using digits ${0, 1, 2, 3, 4, 5}$ such that the number is greater than $4321$.Case $1$: Numbe

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$310$

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(C) We need to form a $4$-digit number $d_1 d_2 d_3 d_4$ using digits ${0, 1, 2, 3, 4, 5}$ such that the number is greater than $4321$.Case $1$: Numbers starting with $5$ $(d_1 = 5)$:There are $6$ choices for each of the remaining $3$ positions $(d_2, d_3, d_4)$.Total $= 1 	imes 6 	imes 6 	imes 6 = 216$.Case $2$: Numbers starting with $4$ $(d_1 = 4)$:Subcase $2.1$: $d_2 > 3$ ($d_2 = 4$ or $5$):There are $2$ choices for $d_2$ and $6$ choices for each of $d_3, d_4$.Total $= 2 	imes 6 	imes 6 = 72$.Subcase $2.2$: $d_2 = 3$:If $d_3 > 2$ $(d_3 = 3, 4, 5)$:There are $3$ choices for $d_3$ and $6$ choices for $d_4$.Total $= 3 	imes 6 = 18$.If $d_3 = 2$:If $d_4 > 1$ $(d_4 = 2, 3, 4, 5)$:There are $4$ choices for $d_4$.Total $= 4$.Summing all cases: $216 + 72 + 18 + 4 = 310$.

The number of four-digit numbers strictly greater than $4321$ that can be formed using the digits $0, 1, 2, 3, 4, 5$ (repetition of digits is allowed) is

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