|
Question 255054: How many different four-digit numbers can be formed using the digits 1, 1, 9, and 9?
Found 2 solutions by CharlesG2, Theo:
Answer by CharlesG2(834)   (Show Source):
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! 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.
|
|
|
| |
