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(B) number greater than $10$ lakh must have at least $7$ digits.Since we have exactly $7$ digits $(2, 3, 0, 3, 4, 2, 3)$,any $7$-digit number formed u

Vedclass · Accepted Answer

$360$

Vedclass · Answer

(B) number greater than $10$ lakh must have at least $7$ digits.Since we have exactly $7$ digits $(2, 3, 0, 3, 4, 2, 3)$,any $7$-digit number formed using these will be greater than $10$ lakh,provided the first digit is not $0$.The total number of permutations of these $7$ digits ($2$ appears twice,$3$ appears thrice,$0$ appears once,$4$ appears once) is given by $\frac{7!}{2! 	imes 3!} = \frac{5040}{2 	imes 6} = 420$.The number of permutations starting with $0$ (which are $6$-digit numbers) is $\frac{6!}{2! 	imes 3!} = \frac{720}{2 	imes 6} = 60$.Therefore,the number of $7$-digit numbers greater than $10$ lakh is $420 - 60 = 360$.

How many numbers greater than $10$ lakh can be formed using the digits $2, 3, 0, 3, 4, 2, 3$?

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