Question 1193951: A binary string is an ordered sequences of 0's and 1's (for example, 0100101). Determine the number of binary strings there are consisting of five 0's and three 1's.
Textbook answer is 56.
Can someone help me solve this problem? Thanks.
Found 2 solutions by Theo, greenestamps:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! your solution here will either be 8C5 of 8C3.
your string is 8 digits long.
8C5 is the number of ways you can get a string of 5 zeroes out of a string of 8 digits that that has 5 zeroes in it. the value of the other 3 digits is inconsequential since they just fill the spaces that don't contain zeroes.
8C3 is the number of ways you can get a string of 3 ones out of a string of 8 digits that has 3 ones in it. the value of the other 5 digits is inconsequential since they just fill the spaces that don't contain ones.
a simple example is a string of 3 digits that contains 2 zeroes and 1 one.
the number of ways of getting 2 zeroes and 1 one would be either 3C2 or 3C1.
3C2 = 3*2/2 = 3
3C1 = 3/1 = 3
there are 3 ways you can get a string of 2 zeroes and 1 one.
they are:
001
010
100
the formula of 8C5 is also shown as C(8,5).
it is equal to 8! / (5! * 3!) which is equal to (8 * 7 * 6 * 5! / (5! * 3!) which is equal to (8 * 7 * 6 / (3 * 2 * 1) which is equal to 56.
it is a combination formula, not a permutation formula.
normally you would use a permutation formula for an ordered set, but in this case, the different sets would be equivalent, as shown below in the simple example.
assume the 3 digits are a,b,c
the number of possible permutations are:
abc
acb
bac
bca
cab
cba
since a and b are equal to 0 and c is equal to 1, this becomes:
001
010
001
010
100
100
3 of the sets are duplicates of the other 3 sets, so the number of possible sets that are different from each other becomes 3.
3P2 is equal to 3! / 1! = 6
3C2 is equal to 3! / (1! * 2!) = 3
3P2 gives you 6 sets where order is important, but since the individual values are not all different, then 3P2 reduces to 3C2.
that's why the combination formula works in this case, where it wouldn't if the individual values were all different from each other.
bottom line:
your answer is either 8C5 or 8C3.
8C5 is the same as C(8,5)
3C3 is the same as C(8,3)
8C5 or C(8,5) is equal to 8! / (5! * 3!)
8C3 or C(8,3) is equal to 8! / (3! * 5!)
as you can calculate, they will both give you the same answer as 56.
let me know if you have any questions.
theo
Answer by greenestamps(13198)  (Show Source):
You can put this solution on YOUR website!
A basic counting principle says that the number of ways of arranging 8 objects in a row of which 5 are one kind and 3 are another is
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