Math Help
- Thread starter hukhalid
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mmm4444bot
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I have no idea what you have learned thus far.
Do any of your lessons talk about something called the Fundamental Counting Principle?
Do you have any specific questions of your own?
Do any of your lessons talk about something called the Fundamental Counting Principle?
Do you have any specific questions of your own?
mmm4444bot
Super Moderator
- Joined
- Oct 6, 2005
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- 10,962
I interpret the assignment to denote that the first digit must be 1, 3, or 6, and the second digit must be 1, 3, or 6, and the third digit must be 1, 3, or 6, and each of the remaining 7 digits must be 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9.
There are 6 digits in total. The numbers with 3 digits, with all digits distinct from each
other, are the permutations of the 6 digits taken 3 at a time, and therefore there are 6*5*4
= 120 of them.
To be odd, one such number must end with 1, 3, or 5. We can construct all of the odd
three digit numbers. For example, the odd numbers ending in 1 are the permutation of the
remaining five digits taken two at a time. So the total number of odd numbers is:
3*(5*4)=60.
A number is larger than 330 if its first digit is 4, 5 or 6, or if its first digit is 3 and its
second digit is 4,5 or 6. In the first case, the number of possibilities is 3 times the number
of permutations of the remaining 5 digits taken 2 at a time, i.e. 3*(5*4)=60. In the second
case, the number of possibilities is 3 times the number of permutations of the remaining 4
digits taken 1 at a time, that is 3*4=12. In total, 72.
regards,
http://www.pioneermathematics.com
other, are the permutations of the 6 digits taken 3 at a time, and therefore there are 6*5*4
= 120 of them.
To be odd, one such number must end with 1, 3, or 5. We can construct all of the odd
three digit numbers. For example, the odd numbers ending in 1 are the permutation of the
remaining five digits taken two at a time. So the total number of odd numbers is:
3*(5*4)=60.
A number is larger than 330 if its first digit is 4, 5 or 6, or if its first digit is 3 and its
second digit is 4,5 or 6. In the first case, the number of possibilities is 3 times the number
of permutations of the remaining 5 digits taken 2 at a time, i.e. 3*(5*4)=60. In the second
case, the number of possibilities is 3 times the number of permutations of the remaining 4
digits taken 1 at a time, that is 3*4=12. In total, 72.
regards,
http://www.pioneermathematics.com
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