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(A) number greater than a million must have at least seven digits. Since we are given exactly seven digits $(2, 3, 0, 3, 4, 2, 3)$,we must use all of

Vedclass · Accepted Answer

$360$

Vedclass · Answer

(A) number greater than a million must have at least seven digits. Since we are given exactly seven digits $(2, 3, 0, 3, 4, 2, 3)$,we must use all of them to form a seven-digit number.The total number of arrangements of these $7$ digits,where $2$ is repeated twice and $3$ is repeated thrice,is given by the formula for permutations of a multiset:$\frac{7!}{2!3!} = \frac{5040}{2 	imes 6} = \frac{5040}{12} = 420$.However,a seven-digit number cannot begin with $0$. If the first digit is $0$,the remaining $6$ digits $(2, 3, 3, 4, 2, 3)$ can be arranged in:$\frac{6!}{2!3!} = \frac{720}{2 	imes 6} = \frac{720}{12} = 60$ ways.Therefore,the number of valid seven-digit numbers is the total arrangements minus those starting with $0$:$420 - 60 = 360$.

How many numbers greater than a million can be formed with the digits $2, 3, 0, 3, 4, 2, 3$?

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