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No. of ways to form a number

Punch

New member
Jan 29, 2012
23
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
 

CaptainBlack

Well-known member
Jan 26, 2012
890
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:

grgrsanjay

New member
Jan 28, 2012
21
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)
 

CaptainBlack

Well-known member
Jan 26, 2012
890
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
 

soroban

Well-known member
Feb 2, 2012
409
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.