BITSAT 2015 Paper

Total number of arrangements of 10 digits 0, 1, 2, ...., 9 by taking 4 at a time =

10

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C4 × 4!
We observe that in every arrangement of 4 selected digits there is just one arrangement in which the digits are in descending order.
∴ Required number of 4-digit numbers =

10 ‌ C4×4!

4!

=

10

‌

C4