Permutations and Combinations Question

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16characterlimit
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#1
Report Thread starter 6 years ago
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http://www.ocr.org.uk/Images/144608-...atistics-1.pdf

Q 4c) and ii), answers are 24 and 54

For 4c, why can't I make each number 1(1) 1(2) 2(1) 2(2) 3(1) 3(2) and use 6x5x4/2!*2!*2! ?

It makes sense to me but it obviously doesn't give the answer, anyone have any good methods to deal with the weirder permutations and combinations questions? Thanks!
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joostan
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(Original post by 16characterlimit)
http://www.ocr.org.uk/Images/144608-...atistics-1.pdf

Q 4c) and ii), answers are 24 and 54

For 4c, why can't I make each number 1(1) 1(2) 2(1) 2(2) 3(1) 3(2) and use 6x5x4/2!*2!*2! ?

It makes sense to me but it obviously doesn't give the answer, anyone have any good methods to deal with the weirder permutations and combinations questions? Thanks!
For c) The ways of including every number at most twice is the same as the total number of ways to make a number minus the ways in which all 3 digits are the same.
This gives the required answer in I think the easiest way.
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(Original post by joostan)
For c) The ways of including every number at most twice is the same as the total number of ways to make a number minus the ways in which all 3 digits are the same.
This gives the required answer in I think the easiest way.
How would I apply this in the second part or in a general example?

The number of all digits being the same or used three times is much more in ii
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joostan
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(Original post by 16characterlimit)
How would I apply this in the second part or in a general example?

The number of all digits being the same or used three times is much more in ii
Well in a 4 digit number, at least one digit must be used twice.
The number of ways of using one digit twice, and the number of ways of using two digits twice are easy enough to calculate.
From there it's a matter of using the possible distinct permutations of each to find the total number.
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