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auniyal
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apache
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Himalayan
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Bluebird
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Using the explanation given by apache above, wouldn't it be a permutation? So, the equation would be 6!/(6-3)! = 6!/3! = 120

This is different than what Himalyan got...do I have the right idea or am I missing something?

Thanks!
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Sergey_is_cool
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Himalayan
auniyal
How many 3 different digit integer can be formed using 6 no's 1,2,3,4,5, 6, if no digit is to be repeated?

Is this permutation or a combination problem? I mean is it 6c3 or 6p3?

= 6x6x6 - 6x5 - 6
= 6 x 31 = 180


I'm getting 6!/(6-3)!=6*5*4=120

how do you get your answer?
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Himalayan
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Sergey_is_cool
Himalayan
auniyal
How many 3 different digit integer can be formed using 6 no's 1,2,3,4,5, 6, if no digit is to be repeated?

Is this permutation or a combination problem? I mean is it 6c3 or 6p3?

= 6x6x6 - 6x5 - 6
= 6 x 31 = 180

I'm getting 6!/(6-3)!=6*5*4=120

how do you get your answer?


agreed. its 6x5x4 = 120

i did the other way counting all same digits and 2 same digits but missed something.

perfect.
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auniyal
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correct me if i am wrong,
okay, out of 6 digits , 3 digits are selected
So, its permutation:- as order is not important, so 6p3 = 6*5*4 = 120

but how did u make sure that digits are not repeated, coz 123 is okay but 112 is not? how was numbers like 112 or 121 truncated from the result.
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apache
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auniyal,
first of all order is important as 123 and 324 are two diffrent digits.

now for your question that how did u make sure that digits are not repeated

How many 3 different digit integer can be formed using 6 no's 1,2,3,4,5, 6, if no digit is to be repeated?

when digits are not repeated =6*5*4 = 6P3
when digits are repeated = 6*6*6

so premutaions are only used when it is assumed that once object is taken it will not be repeated .



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