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GMATPrep Problem .. assign identification

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Re: GMATPrep Problem .. assign identification

by parallel_chase » Thu Sep 11, 2008 8:59 pm
shahab03 wrote:first of all is this permutation or combination problem. also if you can help me understand. thanks



Image

I guess the answer should be 4536.

This problem is of permutations.

We have total 10 letters [0,1,2,3,4,5,6,7,8,9]

first place can be filled in 9 ways, since 0 is not to be included = 9 ways.
Second place can also be filled in 9 ways, since 0 can be included = 9 ways.
Third place can be filled in 8 ways
Fourth place can be filled in 7 ways.

9*9*8*7 = 4536

Whats the OA?
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by msd_2008 » Thu Sep 11, 2008 9:18 pm
How can the answer be 4536?
I would go about this problem in the following way:

1st digit can be arranged in 9 ways (0 not included)
2nd digit can be arranged in 10 ways (0 included)
3rd digit can be arranged in 10 ways (0 included)
4th digit can be arranged in 10 ways (0 included)

Hence total ways = 9*10*10*10 = 9000 ways

We cannot assume that same numbers should not be repeated in different digit places because thats nowhere mentioned in the problem.
If we assume this, then 4536 is the right answer.
Can you please let us know the OA?

Regards
MSD
When the going gets tough, the tough gets going.
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by parallel_chase » Thu Sep 11, 2008 9:22 pm
msd_2008 wrote:How can the answer be 4536?
I would go about this problem in the following way:

1st digit can be arranged in 9 ways (0 not included)
2nd digit can be arranged in 10 ways (0 included)
3rd digit can be arranged in 10 ways (0 included)
4th digit can be arranged in 10 ways (0 included)

Hence total ways = 9*10*10*10 = 9000 ways

We cannot assume that same numbers should not be repeated in different digit places because thats nowhere mentioned in the problem.
If we assume this, then 4536 is the right answer.
Can you please let us know the OA?

Regards
MSD
The question explicitly states that "Each number consists of four different digits"

No Assumptions!!

Hope this helps.
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by cramya » Fri Sep 19, 2008 10:02 pm
Shahab,
This is a permutaion since the order does matter.

Eg: 7586 id is different from 7856 id (even though they have the same digits but the position of the digits make the difference)

Make 4 slots

1st slot can be filled by 9 numbers(since 0 cannot be included)
2nd slot by 9(the first number u have choosen cannot be included but 0 can be the 2nd digit)
3rd slot can be filled by 8 since we cant use the first 2 numbers in the first 2 slots
4th slot by 7 (same reason as above)

So 9*9*8*7 would be 4536
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by cramya » Fri Sep 19, 2008 10:10 pm
Easier way to understand this would be

Take 0,1,2 How many different 3 digit identifcation numbers can be formed where 0 cannot be the first digit(since its 012 is not valid)

0,1,2

Brute force listing of possibilities:
102
120
201
210

4 numbers


Permutation/counting principle method:

1st slot: 2
2nd slot: 2
3rd slot:1

2*2*1 = 4

Hope this helps!
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