From 0 to 9, four digited numbers can be formed such that the digits are in ascending order is

From 0 to 9, four digited numbers can be formed such that the digits are in ascending order is

A
$^{10} P_{4}$
B
$^{10} C_{4}$
C
$^{10} P_{4} -^{9} P_{3}$
D
$^{10} C_{4} -^{9} C_{3}$

Solution:

The number of 4 digited numbers formed from 0 to 9 in a specified order = $\frac{P (10, 4)}{4!} = C (10, 4)$

If 0 occupies the thousand's place , then the number of 4 digited numbers in a specified order = C( 9,3)

$∴$ No of 4 digit numbers from 0 to 9 number of 3 digit numbers beginning with 0 in specified order is $^{10} c_{4} -^{9} c_{3}$

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