(C) number lying between $1000$ and $10000$ is a $4$-digit number.We have $7$ distinct digits available: ${1, 3, 5, 6, 7, 8, 9}$.Since no digits are r

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(C) number lying between $1000$ and $10000$ is a $4$-digit number.We have $7$ distinct digits available: ${1, 3, 5, 6, 7, 8, 9}$.Since no digits are repeated,we need to arrange $4$ digits out of the $7$ available digits.The number of ways to do this is given by the permutation formula $^nP_r = \frac{n!}{(n-r)!}$.Here,$n = 7$ and $r = 4$.Therefore,the number of ways is $^7P_4 = 7 \times 6 \times 5 \times 4 = 840$.

Competitive Quantitative Aptitude Permutation and Combination Permutation and Combination

Medium

How many numbers lying between $1000$ and $10000$ can be formed by using the digits $1, 3, 5, 6, 7, 8, 9$ if no digits are repeated?

A
$940$
B
$640$
C
$840$
D
$740$

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How many numbers lying between $1000$ and $10000$ can be formed by using the digits $1, 3, 5, 6, 7, 8, 9$ if no digits are repeated?

Similar Questions

In how many ways can five examination papers be arranged so that physics and chemistry papers never come together?

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