Arrangements of binary numbers

Toast · 2006-12-25 03:33:08

Question:
A binary number consists of 1's and 0's. How many different binary numbers can be made using four 1's and two 0's.
(A binary number cannot start with 0.)
(Answer = 10)

Working so far:
Well, in total there are

=15 ways of arranging it if it CAN start with a 0. I don't know how to count the number of combinations with 1 in front of it without writing out each and every combination starting with 0 and subtracting that from the total. Help please?

Sorry for all these counting questions lol, I'll be finished with this chapter soon.

Ricky · 2006-12-25 04:07:49

Well, we need a 1 to go first, so we have:

1 _ _ _ _ _

Out of these 5 slots, we need to choose 2 for the 0's, which is 5 choose 2 or C(5, 2). Now there are three 1's left, so we need to put these into the three slots, which is 3 choose 3 or rather, just 1.

So the final answer is 5 choose 2 = 10.

Don't be afraid of using combinations. They are your friends, more often than not.

Toast · 2006-12-25 04:48:19

What do you mean by 5 choose 2? I haven't really learnt much about arranging indistinguishable objects with restriction.

Ricky · 2006-12-25 08:27:23

If you have n "slots" (places where objects can go), and you wish to know how many places r identical objects can go, then the general formula is known as a combination:

pi man · 2006-12-25 18:10:30

I like combinations also. But it can be done your way also. You had 6! / (4! * 2!) but you didn't know how to exclude the numbers that began with a 0. Well the easiest way to do that is to not include them to begin with. The valid numbers have to begin with a 1 so that leaves three 1's and two 0's to arrange. You know how to do that. There's 5 digits so you start on with 5!. Divide by 3! because the 1 occurs 3 times. And then divide by 2! because the 0 occurs twice. You now have 5! / (3! * 2!) = 120 / (6*2) = 120/12 = 10.

mathsyperson · 2006-12-26 02:06:06

Alternatively, we could use that you've already worked out that if the first digit has no restrictions on it, then there would be 15 possible combinations. As there are two 0's out of the six numbers, that means that for any of those combinations, there is a 2/6 chance that the first number will be 0.

We only want the numbers that don't start with 0, so we multiply the 15 by 4/6 and get 10, the right answer.

I like Ricky's way better, because it's less confusing, but it's another way of thinking about it anyway.

Toast · 2006-12-26 20:01:51

Thanks everyone, I finally understand. I think I will use pi man's method because although I am sure ricky and mathsyperson's methods are useful too in their own ways, the textbook examples are most similar to his.

krassi_holmz · 2006-12-27 08:59:30

Just an obsrvation - because the numbers are in base 2, the total amount of the numbers, starting with 1 is equal to the total amount of numbers, starting with 0. But we can't use this in this problem, because the number of the nulls is fixed.

Math Is Fun Forum

#1 2006-12-25 03:33:08

Arrangements of binary numbers

#2 2006-12-25 04:07:49

Re: Arrangements of binary numbers

#3 2006-12-25 04:48:19

Re: Arrangements of binary numbers

#4 2006-12-25 08:27:23

Re: Arrangements of binary numbers

#5 2006-12-25 18:10:30

Re: Arrangements of binary numbers

#6 2006-12-26 02:06:06

Re: Arrangements of binary numbers

#7 2006-12-26 20:01:51

Re: Arrangements of binary numbers

#8 2006-12-27 08:59:30

Re: Arrangements of binary numbers

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