Question 1094683: A four-digit number is formed using four of the eight digits 1, 2, 3, 4, 5, 6, 7, 8. No digit can be used more than once. Find how many different four-digit numbers can be formed if the number is odd and greater than 6000.
So, I've tried working it out several times, but my answer isn't the same as the textbook's answer. My answer is 3*6*5*4 = 360, while the textbook says it is 330. Which one is correct?
Answer by ikleyn(52781)   (Show Source):
You can put this solution on YOUR website! .
A four-digit number is formed using four of the eight digits 1, 2, 3, 4, 5, 6, 7, 8. No digit can be used more than once.
Find how many different four-digit numbers can be formed if the number is odd and greater than 6000.
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I promised to complete this problem.
Now it is good time for me to do it. Sorry for the delay.
Let's solve it in three steps.
First consider the numbers starting with "6" as leftmost digit.
Then consider the numbers starting with "8" as leftmost digit.
Finally, will consider the numbers starting with "7" as leftmost digit.
1. Consider the numbers starting with "6" as leftmost digit.
You may (and should) use any of four odd digits at the end: 1, 3, 5 and 7.
It gives you 4 choices.
For the second and third digit you can use any of remaining 8-1-1 = 6 digits.
In all, it gives you 4*6*5 = 120 choices/numbers.
2. Consider the numbers starting with "6" as leftmost digit.
The same logic and the same analysis gives you 120 other choices/numbers.
So far, so good, and you just have 120 + 120 = 240 numbers.
3. Consider the numbers starting with "7" as leftmost digit.
You may (and should) use any of remaining three odd digits at the end: 1, 3, and 5.
It gives you 3 choices.
For the second and third digit you can use any of remaining 8-1-1 = 6 digits.
In all, it gives you 3*6*5 = 90 choices/numbers.
4. Adding everything together, you have 120 + 120 + 90 = 330 numbers/choices,
exactly as in your textbook.
H a p p y l e a r n i n g ! !
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