Number Property, High Difficulty Level Problem

[beboppin]

Originally Posted by CyberSpy

Element, instead of considering 2 digit nos and 3 digit nos seperately this problem can simply be solved by considering XXXX and filling these four places with the required digits. I am able to wok this out as follows,

1,4

4x3 = 12 possible values.

Similarly for 3,2 12 values
Those nos with zeros preceding them are the reqd 2 digit and 3 digit nos. you dont have to calculate them seperately. All you have to do id to do away with the subtraction part.



For 1,2,2

1220
1202
1022
2120
0122
0212
2102
2012
2210
2021
2201
0221

i.e 4x3= 12(4 positions for 1 and 3 postions for the rest of the numbers for each positionsof 1).Can someone give me a better explanation of this? Im not satisfied with mine.
Similar 12 combinations for 1,1,3

4 combinations each for 1,1,1,2 and 0,0,0,5

Adding these we get 56.(12+12+12+12+4+4).

For 2, 2, 1, I would say that you can do 4 positions for one number, 3 positions for the second, and two positions for the third. Because two numbers repeat (i.e. 2 and 2), you'd need to divide by 2!, or 2, so you end up at 4*3

[CyberSpy]

Originally Posted by beboppin

For 2, 2, 1, I would say that you can do 4 positions for one number, 3 positions for the second, and two positions for the third. Because two numbers repeat (i.e. 2 and 2), you'd need to divide by 2!, or 2, so you end up at 4*3

To put it in a better way, it is this:


combinations for no 1 x combinations for no 2 X 1 (for 0)

4C1 X 3C2

= 4x 3!/2!..

or

4C2 X 2C1

4!/2! x 2 = 12

[MBAchase]

Yes 56 is the correct answer.

[kssshah]

I have just started my studies for GMAT and so this may be a trivial question. But please help me understand why this is a combination problem and not a permutation problem considering the order in which the #s come up matter. For e.g. 1220 is not the same as 1022. Combination of the #s here is the same but the order is different which, I would imagine, is a permutation problem.