Maths Question....................
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Maths Question....................

[From: ] [author: ] [Date: 11-10-02] [Hit: ]
The 100th position can have 4 options since we have 4 numbers ( 1 , 2 , 3 ,The 1th position can only have 2 choices.123 124 132 134 142 143 212 213 214 231 234 241 243 312 314 321 324 341 342 412 413 421 423 431 432-This is simple: 4 possible digits for place #1, 3 possible digits for place #2,......
How many 3 digit numbers can be written using the digits 1,2,3,4? Each digit can only be used once in any number

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You should permutations and combinations here

XXX thats the numbers possible. Now the condition is each digit can be only once , so the approach should be this

The 100th position can have 4 options since we have 4 numbers ( 1 , 2 , 3 , 4 )
The 10th position can now have 3 options since we only have now 3 numbers left now to work with due to the restriction in our Q here
The 1th position can only have 2 choices. Now the we will have to multiply those so
4 * 3 * 2 = 24; Lets see what all are those
123 124 132 134 142 143 212 213 214 231 234 241 243 312 314 321 324 341 342 412 413 421 423 431 432

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This is simple: 4 possible digits for place #1, 3 possible digits for place #2, because 1 digit was used already, and 2 possible digits for place #3, since 2 were used already. That means the answer is 4*3*2 = 24 possible combinations.

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Hello Joya,
I ran the four digits, 1..2..3..4 through my gridwork, and the largest number of three-digit numbers I came up with is 19...........

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16
1
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