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#1 2006-12-21 17:16:38
- balasvce
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- Registered: 2006-12-20
- Posts: 37
jutify please
31) how many 4 digit numbers possible of format xyyx where
x!=0 and
x!=y (ans:9*9=81)
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#2 2006-12-22 02:53:46
- mathsyperson
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- Registered: 2005-06-22
- Posts: 4,900
Re: jutify please
I got really confused there for a while, thinking you were talking about factorials. By !=, you mean ≠, right?
In that case, the answer's right.
You're told that x ≠ 0, which means that it can be any of 1, 2, 3, 4, 5, 6, 7, 8 or 9 (9 possibilities).
y can be any of the above possibilities excluding the value that x took, but it can also be 0, which means that there are also 9 possibilities for y.
Therefore, the total number of possibilities is 9*9 = 81.
Why did the vector cross the road?
It wanted to be normal.
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#3 2006-12-22 07:35:15
- John E. Franklin
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- Registered: 2005-08-29
- Posts: 3,588
Re: jutify please
What if y is zero. Then you get 9*10 or 90 ways.
igloo myrtilles fourmis
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#4 2006-12-22 07:45:40
- Ricky
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- Registered: 2005-12-04
- Posts: 3,791
Re: jutify please
John, you forget that x can not be the same as y. So lets say x is 5. Then y can be 0, 1, 2, 3, 4, 6, 7, 8, 9, which is 9 possibilities.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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