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- Jan 26, 2012

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The side conditions mean that the leading digit must be one of 1,2,3,4,5, and the last digit 1,3,5,6,9.A six-digit number is to be formed from the digits 1, 2 ,3, 4, 5, 6, 7, 8, 9. Find how many ways the six-digit number can be formed if the number must be odd and is less than 600,000 and no digit may appear more than once

Decompose into two groups, those with an odd leading digit and those with an even leading digit.

CB

Last edited:

- Jan 28, 2012

- 21

1st digit can be 1,2,3,4,5

2nd digit can be 0-9

3rd digit can be 0-9

4th digit can be 0-9

5th digit can be 0-9

6th digit can be 1,3,5,7,9

total ways = 5(10)(10)(10)(10)(5)

- Jan 26, 2012

- 890

No, as you choose the digits the number of options for the subsequent digits goes down.

1st digit can be 1,2,3,4,5

2nd digit can be 0-9

3rd digit can be 0-9

4th digit can be 0-9

5th digit can be 0-9

6th digit can be 1,3,5,7,9

total ways = 5(10)(10)(10)(10)(5)

CB

A six-digit number is to be formed from the digits 1, 2 ,3, 4, 5, 6, 7, 8, 9.

Find how many ways the six-digit number can be formed if the number must be odd

and is less than 600,000 and no digit may appear more than once.

The number is odd; the last digit must be 1, 3, 5, 7 or 9.

The number is less than 600,00; the first digit must 1, 2, 3, 4 or 5.

There are two cases to consider.

(1) The last digit is 1, 3 or 5: .$3$ choices.

. . .The first digit has only $4$ choices.

. . .There are $\text{P}(7,4)$ ways to arrange the other four digits.

There are: .$ 3\cdot4\cdot\text{P}(7,4) \,=\,10,080 $ ways.

(2) The last digit is 7 or 9: .$2$ choices.

. . .The first digit has all $5$ choices.

. . .There are $\text{P}(7,4)$ ways to arrange the other four digits.

There are: .$ 2\cdot5\cdot\text{P}(7,4) \,=\,8,400 $ ways.

Therefore, there are: .$10,080 + 8,400 \:=\:18,480 $ ways.