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

30

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60

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45

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90

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Solution

The correct option is A $60$
$- - - - - - - - 3, 4, 7, 5, 5$
No. of five digit numbers $= \frac{5!}{2}$
$= \frac{120}{2} = 60$

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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

The correct option is A 60−−−−−−−−3,4,7,5,5No. of five digit numbers=5!2=1202=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

The correct option is A $60$
$- - - - - - - - 3, 4, 7, 5, 5$
No. of five digit numbers $= \frac{5!}{2}$
$= \frac{120}{2} = 60$