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 January 29th, 2012, 12:54 AM #1 Member   Joined: Jan 2012 Posts: 72 Thanks: 0 No. of ways to form a number 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
January 29th, 2012, 03:37 AM   #2
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Re: No. of ways to form a number

Quote:
 Originally Posted by 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
....no digit may appear more than once?

123457, 123459, ...., 598761, 598763 ; 18480

January 29th, 2012, 04:35 AM   #3
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Re: No. of ways to form a number

Hello, Punch!

Quote:
 A six-digit number is to be formed from the digits 1, 2 ,3, 4, 5, 6, 7, 8, 9. Find how many ways a 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.

$\text{W\!e have a 6-digit number: }\:N \:=\:a\,b\,c\,d\,e\,f$

$\begin{array}{ccc}N\,<\,600,000: && a\text{ will be: }\,\{1,\,2,\,3,\,4,\,5\} \\ \\ \\ N\text{ is odd:} && f\text{ will be: }\,\{1,\,3,\,5,\,7,\,9\} \end{array}$

There are two cases to consider.

$(1)\:f\,=\,\{1,\,3,\,5\}\;\text{ . . . 3 choices}$

$\text{Then }a\text{ has 4 choices.}$

$\text{There are: }\,_7P_4\text{ ways to choose and place the other 4 digits.}$

$\text{Hence, there are: }\:3\,\cdot\,4\,\cdot\,_7P_4 \:=\:10,080\,\text{ such numbers.}$

$(2)\:f\,=\,\{7,\, 9\}\:\text{ . . . 2 choices}$

$\text{Then }a\text{ has 5 choices.}$

$\text{There are: }\,_7P_4\text{ ways to choose and place the other 4 digits.}$

$\text{Hence, there are: }\:2\,\cdot\,5\,\cdot\,_7P_4 \:=\:8,400\,\text{ such numbers.}$

$\text{Therefore, there are: }\:10,080\,+\,8,400 \:=\:18,480\text{ odd numbers less than 600,000.}$

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