Permutaions

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Permutaions

by sudhir3127 » Mon Jul 21, 2008 2:05 am
Need your help on this

1. How many numbers between 50000 and 60000 can be formed using the digits 2 to 7 when any digit can occur any number of times.

a. 1296
b. 625
c. 7776
d. 2520

OA is A.
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Re: Permutaions

by Ian Stewart » Mon Jul 21, 2008 5:06 am
sudhir3127 wrote:Need your help on this

1. How many numbers between 50000 and 60000 can be formed using the digits 2 to 7 when any digit can occur any number of times.

a. 1296
b. 625
c. 7776
d. 2520

OA is A.
The first digit must be a 5: one choice
The four other digits can be any number from 2 to 7: six choices for each.

Multiply the number of choices: 1*6*6*6*6 = 1296.
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by sudhir3127 » Mon Jul 21, 2008 7:53 am
thanks Ian for your reply....

was just wondering how did u get 6 ..if we have already used 5 for the first position ..then we are left with only 2,3,4,6,7 choosing any one from these shud be 5 ..

please let me know where i am wrong?

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by parallel_chase » Mon Jul 21, 2008 10:08 am
sudhir3127 wrote:thanks Ian for your reply....

was just wondering how did u get 6 ..if we have already used 5 for the first position ..then we are left with only 2,3,4,6,7 choosing any one from these shud be 5 ..

please let me know where i am wrong?
The question stem states that the numbers can be repeated. Therefore you have 2, 3, 4, 5, 6, 7. The first number is 5 because the condition states that the number has to between 50000 and 60000.

If the question would have stated no repetition is allowed, then the function would have looked like this

1*5*4*3*2

Let me know if you still have any doubts.

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by sudhir3127 » Mon Jul 21, 2008 10:14 am
i dint get that combination of 1*5*4*3*2 ....

even i have doubt in the other question which i posted and u replied with the same type of combination.... can u please explain how did u get this combination?

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by parallel_chase » Mon Jul 21, 2008 11:03 am
Alrite, I think you have a problem with permutations and combination. Nothing to worry about you can master them in less an hour.

Here is a fast track lesson in one post.

First things first forget about factorial and formulas for permutation and combination. Keep them aside, if you really understood the whole concept you wont be needing any formulas.

Always remember:

Simultaneous actions always MULTIPLY keyword = "and"
Actions occurring not simultaneously always ADD keyword="or"

There are 3 bowls and 5 balls. In how many ways can you put 5 balls in 3 bowls. Assume each bowl can hold only 1 ball.

You can put 5 balls in 1st bowl in 5 ways
You can put 4 balls in 2st bowl in 4 ways (since one ball has already been put in 1st bowl and now you are left with 4 balls)

Similarly, you can put 3 balls in 3rd bowl in 3 ways

Since all this has to happen simultaneously,
5*4*3 = 60 ways

Now lets look at the formula 5p3 = 60 ways.

Now switch the wordings of the question.

How many 3 digit numbers can formed using letters 1, 2, 3, 4, 5. None of the digits can be repeated.

Answer is again 60.

In GMAT the question regarding digits wont be this easy. There will be some conditions, like you saw in previous questions where divisibility and other things are taken into account. I think you are pretty good at number system so you wouldnt have any problems dealing with those.

Finally, I'll answer your question about 1*5*4*3.

We have to find numbers between 50000-60000 and we have 2,3,4,5,6,7 digits. The numbers can be repeated.

_ _ _ _ _ _

In the first blank we can only put 1 number i.e. 5 because the number formed has to be between 50000 to 60000. Therefore we can arrange this only 1 ways.

Second blank we can put any number from 2-7 since the question says the number can be repeated. Therefore we have 6 options = 6 ways.

Similarly we have 6 letters for each of the 5 blanks except for the first one.
And all this has to happen simultaneously therefore 1*6*6*6*6*6

Try to do the same problem with condition that numbers cannot be repeated.


This is the best that I can do in one post. Hope this helps you, it indeed has helped me.