MHB How Many Ways to Form a Six-Digit Odd Number Less Than 600,000?

  • Thread starter Thread starter Punch
  • Start date Start date
  • Tags Tags
    Form
AI Thread Summary
To form a six-digit odd number less than 600,000 using the digits 1-9 without repetition, the first digit must be 1-5 and the last digit must be 1, 3, 5, 7, or 9. Two cases are analyzed: when the last digit is 1, 3, or 5, there are 10,080 combinations; when the last digit is 7 or 9, there are 8,400 combinations. The total number of valid combinations is calculated to be 18,480. This approach ensures all conditions are met while accounting for the decreasing options as digits are chosen. The final result confirms the total ways to form the specified six-digit number.
Punch
Messages
44
Reaction score
0
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
 
Physics news on Phys.org
Punch said:
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 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.

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

CB
 
Last edited:
its a six digit number
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)
 
grgrsanjay said:
its a six digit number
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)

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

CB
 
Hello, Punch!

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.
 
Back
Top