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

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In summary, there are 18,480 ways to form a six-digit number using the digits 1, 2 ,3, 4, 5, 6, 7, 8, 9 that is odd, less than 600,000, and with no repeating digits.
  • #1
Punch
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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
 
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  • #2
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
 
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  • #3
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)
 
  • #4
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
 
  • #5
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.
 

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

What is meant by "No. of ways to form a number"?

"No. of ways to form a number" refers to the number of unique combinations or arrangements in which a specific number can be formed using a set of digits or elements.

How is the number of ways to form a number calculated?

The number of ways to form a number can be calculated using the formula nPr = n!/(n-r)!, where n is the total number of elements and r is the number of elements used to form the number.

Can the number of ways to form a number be greater than the number itself?

Yes, the number of ways to form a number can be greater than the number itself. This occurs when there are repeated digits or elements in the set, allowing for multiple combinations to form the same number.

How does the order of the digits or elements affect the number of ways to form a number?

The order of the digits or elements does affect the number of ways to form a number. Different arrangements of the same set of elements can result in different numbers and therefore, different ways of forming the number.

Can the number of ways to form a number be negative?

No, the number of ways to form a number cannot be negative. It is always a positive integer, as it represents the total number of unique combinations or arrangements possible.

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