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#1 2006-12-21 17:16:38

balasvce
Member
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
Moderator
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
Member
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
Moderator
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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