# number of permutation of 0,1,....,9 such that odd number are increasing order?

### 2 Answers

- Φ² = Φ+1Lv 79 months ago
There are 10! arrangements of these digits.

There are 5! arrangements of 1, 3, 5, 7, 9, only one of which has them in this precise order.

The number of permutations is then 10!/5! = 30240.

There is no need to be concerned with the leading 0 as we are discussing the digits as characters. We might just as well have been asked about 5 vowels dispersed amongst another 5 consonants.

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- 冷眼旁觀Lv 69 months ago
Firstly, arrange 1,3,5,7,9 in order. Number of permutation = 1

_1_3_5_7_9_ (There are 6 spaces as shown.)

Then, put "2" in one of the 6 spaces above. Number of permutation = 6

_X_X_X_X_X_X_ (There are 7 spaces as shown.)

Then, put "4" in one of the 7 spaces above. Number of permutation = 7

_X_X_X_X_X_X_X_ (There are 8 spaces as shown.)

Then, put "6" in one of the 8 spaces above. Number of permutation = 8

_X_X_X_X_X_X_X_X_ (There are 9 spaces as shown.)

Then, put "8" in one of the 9 spaces above. Number of permutation = 9

_X_X_X_X_X_X_X_X_ (There are 10 spaces as shown.)

Lastly, put "0" in one of the 10 spaces above. Number of permutation = 10

_X_X_X_X_X_X_X_X_X_ (There are 10 spaces as shown.)

Total number of permutation

= 1 × 6 × 7 × 8 × 9 × 10

= 30240

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Special case:

If "0" is NOT allowed to put in front of the other number, number of permutation for "0" = 9

Then, total number permutation

= 1 × 6 × 7 × 8 × 9 × 9

=27216

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