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Intern
Joined: 18 Dec 2012
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(1) The Sum of all 4 digit nos(without repetition)using digits 1,2,3,4,5
(2) The Sum of all 4 digit nos(with repetition)using digits 1,2,3,4,5
(3) The Sum of all 4 digit nos(without repetition)using digits 1,2,2,3,4,5,5
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BSchool Thread Master
Joined: 23 Mar 2011
Posts: 593
Concentration: Healthcare, Strategy
GMAT 1: 720 Q51 V35
GMAT 2: 770 Q51 V44
WE: Consulting (Health Care)
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249
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Intern
Joined: 21 Dec 2012
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(1)
Another way to approach this problem is to recognize that the way the sequence increases from the min (1234) is symmetrical to the way it decreases from the max (5432). Therefore if you find the average of the min and max and multiply it by the number of possibilities (5*4*3*2 or 120) then you'll have your answer.
\frac{1234+5432}{2} = 3333
3333*120 = 399960
(2)
You can use the same method here. The min is 1111 and the max is 5555. The number of possibilities is 5*5*5*5 = 625
\frac{1111+5555}{2} = 3333
3333*625 = 2083125
(3)
For this one we need to be a little more clear. Would the sum include duplicate numbers or should all numbers be distinct? For example, the number 1223 can be formed two ways. The first way would be 1 (1st 2) (2nd 2) 3. The other way would be 1 (2nd 2) (1st 2) 3. If all numbers need to be distinct then the solution will be difficult. I used excel to find the sum to be 933240. I would be very interested if someone knows how to do this problem assuming all numbers must be distinct. Anyways, the method I used above won't work since you lose symmetry. Instead you can evaluate using Bunuel's solution like so:
There are seven numbers that we can put into a group of four: 1, 2, 2, 3, 4, 5, 5
The number of possibilities is 7*6*5*4 = 840
840/7 = 120
Therefore the sum = 1111*(120*(1+2*2+3+4+2*5)) = 2933040
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