How many 5 digit nos are there if the 2 leftmost digits are

Question

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?

mikemcgarry · Answer

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 24300 . 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

Bunuel · Answer

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.

KingsAgain · Answer

Different approach:

If the number must be  O O # # # , then there are 5 * 5 * 10 * 10 * 10 = 25,000 total possibilities

There are four different cases that can result the number 4 being used more than once:
A. O O 4 4 4 = 5 * 5 = 25
B. O O 4 # 4 = 5 * 5 * 9 = 225
C. O O 4 4 # = 5 * 5 * 9 = 225
D. O O # 4 4 = 5 * 5 * 9 = 225

Total possibilities = 25,000 - 25 - 225 - 225 - 225 = 24,300

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How many 5 digit nos are there if the 2 leftmost digits are

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How many 5 digit nos are there if the 2 leftmost digits are

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