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The addition problem above shows four of the 24 different

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The addition problem above shows four of the 24 different [#permalink] New post 03 Dec 2005, 17:30
The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,3 & 4 exactly once in each integer. what is the sum of these 24 integers ?
A. 24000
B. 26664
C. 40440
D. 60000
E. 66660

How come there are only 24 possible combinations ? when I think there could be only 16 ??

Is it a combination questions ???
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Re: Maths OG 176 Additional problem [#permalink] New post 08 Dec 2005, 04:44
briozeal wrote:
The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,3 & 4 exactly once in each integer. what is the sum of these 24 integers ?
A. 24000
B. 26664
C. 40440
D. 60000
E. 66660

How come there are only 24 possible combinations ? when I think there could be only 16 ??

Is it a combination questions ???


I don't how to call this type of question
But some combo definitely using here
There is 4! ways to place 4 integers 4!=24. Hence there is 24 different intgers consisting of 4 digits!
1234 2134
1342 2341
1423 2431
1243 2143
1324
1243
And so on!But you don't need to draw every possible combination of digits
You need only to remember that 4 digits may be arranged in 4!ways which is equal to 4*3*2*1=24 .For Example 5 digits may be arranged 5! ways 5!=5*4*3*2*1=120 Hope his helps!

By the way the answer Is E!I think
Later will explain if you need
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 [#permalink] New post 08 Dec 2005, 08:22
First digit = 1 or 2 or 3 or 4
Second digit = 1 or 2 or 3 or 4 (= 10 or 20 or 30 or 40)
Third digit = 1 or 2 or 3 or 4 (= 100 or 200 or 300 or 400)

There are 24 ways of making a three digit number, and this means that each number is used 6 times in each digit.

100 * 6 = 600
200 * 6 = 1200
300 * 6 = 1800
400 * 6 = 2400
-----------------
Sum = 6000

10 * 6 = 60
20 * 6 = 120
30 * 6 = 180
40 * 6 = 240
-----------------
Sum = 600

1 * 6 = 6
2 * 6 = 12
3 * 6 = 18
4 * 6 = 24
----------------
Sum = 60

Total sum = 6660

What am I missing here?
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 [#permalink] New post 08 Dec 2005, 08:28
gamjatang wrote:
First digit = 1 or 2 or 3 or 4
Second digit = 1 or 2 or 3 or 4 (= 10 or 20 or 30 or 40)
Third digit = 1 or 2 or 3 or 4 (= 100 or 200 or 300 or 400)

There are 24 ways of making a three digit number, and this means that each number is used 6 times in each digit.

100 * 6 = 600
200 * 6 = 1200
300 * 6 = 1800
400 * 6 = 2400
-----------------
Sum = 6000

10 * 6 = 60
20 * 6 = 120
30 * 6 = 180
40 * 6 = 240
-----------------
Sum = 600

1 * 6 = 6
2 * 6 = 12
3 * 6 = 18
4 * 6 = 24
----------------
Sum = 60

Total sum = 6660

What am I missing here?


I forgot to consider the fourth digit... :oops:

The answer should be (E).

(1111*6)+(2222*6)+(3333*6)+(4444*6) = 66660
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Auge um Auge, Zahn um Zahn :twisted: !

  [#permalink] 08 Dec 2005, 08:28
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