First slide

Permutations

Question

The number of five digit numbers that contain 7 exactly once is

Moderate

A

$(41) (9^{3})$
B

$(37) (9^{3})$
C

$(7) (9^{4})$
D

$(41) (9^{4})$

Solution

When 7 is right in the beginning, the number of numbers $= 9^{4}$ and when 7 is not in the beginning, the number of numbers is $({}^{4}C_{1}) (8) (9^{3})$
$∴$ Required number of numbers
$= 9^{4} + (^{4} C_{1}) (8) 9^{3} = (41) (9^{3})$

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