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02 Feb 2012, 04:59
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How many 5 digit nos are there if the 2 leftmost digits are odd and the digit 4 can't appear more than once in the number ?
Could someone please provide a solution using a approach other than ( 1-x(none)) approach?
Could someone please provide a solution using a approach other than ( 1-x(none)) approach?
02 Feb 2012, 15:12
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Hi, there. I'm happy to help with this. 
It's hard to do proper indenting in the plain text format of these posts, so I prepared a pdf addressing this question.
The numerical answer I got was. According to my calculations, that's how many five-digit numbers meet these conditions.
Does everything I've said make sense? Please do not hesitate to ask if you have any questions.
Mike
It's hard to do proper indenting in the plain text format of these posts, so I prepared a pdf addressing this question.
The numerical answer I got was
24300
Does everything I've said make sense? Please do not hesitate to ask if you have any questions.
Mike
Attachments
five digit numbers, two odd digits, 4 only once.pdf [65.49 KiB]
Downloaded 378 times
02 Feb 2012, 15:35
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abhi47
A shorter way of the same concept as Mike used.
There can be two cases:
A. 4 is used once: OO**4 --> (5*5*9*9)*3: 5 choices for the first digit as there are 5 odd numbers, 5 choices for the second digit for the same reason, 9 choices for one of the two * (not-4 digit), 9 choices for another * (not-4 digit), multiplied by 3 as 4 can take the place of any of the three last digits (OO**4, OO*4*, OO4**);
B. 4 is not used: OO*** --> 5*5*9*9*9: the same logic as above.
5*5*9*9*3+5*5*9*9*9=24,300.
