Question 255054
should be equal to (4!)/(2!*2!) = (4*3*2*1)/(2*1*2*1) = 6


let's see what they would be:


1199
9911
1919
9191
1991
9119


this winds up being a permutation with some of the elements being the same.


formula for a permutation is n!


if x of the n are the same, then the formula becomes n!/x!


if, in addition to x being the same, y are also the same, then the formula becomes n!/(x!*y!)


in your problem, n was equal to 4.
x was equal to 2 because you had 2 ones.
y was also equal to 2 because you had 2 nines.


n!/(x!*y^1) became 4!/(2!*2!)


to see what the difference is, look at 3 letters (a,b,c)


how many ways can they be formed?


3! = 3*2*1 = 6 ways because each of the letters is different.


they are:


abc
acb
bac
bca
cab
cba


take the same 3 letters and make 2 of them the same.


let's assume a and b become the same letter d.


you would then have ddc


how many ways can they be formed.


we have n = 3 and x = 2 and our formula is n!/x! = 3!/2! = 3


the number of ways they can be formed is 3 as follows:


ddc
dcd
cdd


please keep in mind that we are talking about permutations here, and not combinations.


with permutations, order is important.   the same elements in a different order are a different set.


with combinations, order is not important.   each set has to have different elements in it.


take the set of abc.


with combinations, this forms one set.   the same 3 elements can only be used once regardless of what order they are in the set.


with permutations, this forms 6 sets.  the same 3 elements can be used 6 times in a different order each time.


the number of combinations possible would be 1.
the numer of permutations possible would be 6.