How many numbers consisting of 5 digits can b | vedclass

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(B) To form a $5$-digit number using the digits $3, 4, 7$ (each once) and $5$ (twice),we need to arrange the set of digits ${3, 4, 7, 5, 5}$.The total

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$60$

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(B) To form a $5$-digit number using the digits $3, 4, 7$ (each once) and $5$ (twice),we need to arrange the set of digits ${3, 4, 7, 5, 5}$.The total number of arrangements of $n$ objects where $n_1$ objects are of one kind,$n_2$ are of another kind,etc.,is given by the formula $\frac{n!}{n_1! n_2! \dots}$.Here,the total number of digits $n = 5$.The digit $5$ repeats $2$ times,so $n_1 = 2$.The digits $3, 4,$ and $7$ appear $1$ time each.Therefore,the number of ways is $\frac{5!}{2!} = \frac{120}{2} = 60$.

How many numbers consisting of $5$ digits can be formed in which the digits $3, 4$ and $7$ are used only once and the digit $5$ is used twice?

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