Math Is Fun Forum

LuisRodg

Real Member

Registered: 2007-10-23

Posts: 322

How many bit strings of length 10:

(d) equal number of 0s and 1s.

Since its length 10 and its equal numbers then the string has to contain 5 0s and 5 1s. I tried to think of it as one group of 0s and one group of 1s. The way you can arrange the group of 0s is 5! and the way you can arrange the group of 1s is 5! as well. You can also flip this two groups so I thought my answer was:

2 * 5! * 5! = 480

but its wrong.

How do you think of this?

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A coin is flipped 10 times. How many possible outcomes?

(a) in total
(b) exactly two heads
(c) at most three tails
(d) contain the same number of tails and heads.

I found (a) to be 2^10 but I have no idea how to do the rest?

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How many subsets with more than 2 elements does a set with 100 elements have?

A set with 100 elements has 2^100 subsets. But how do I find how many subsets with more than 2 elements?

I tried to think of it like this, total subsets is 2^100 so if I want how many subsets with more than 2 elements then it would be:

2^100 - (subsets with no elements) - (subsets with 1 element) - (subsets with 2 elements)
2^100 - 2^0 - 2^1 - 2^2
2^100 - 1 - 2 - 4
2^100 - 7

but its wrong. Could anyone help me with all this questions?

Last edited by LuisRodg (2008-04-01 05:53:37)

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: How many bit strings of length 10:

d.)  5  heads and 5 tails, every permutation:

The internet says do 10!/5!/5! = 252.

252 = 5 + 5 + 5 + 5 + 40 + 40 + 40 + 40 + 36 + 36 (my notes)

h = heads
t = tails
g = gap to put the other kind in; they alternates between h's and t's.

1:
h

2:
hh or hgh

3:
hhh hghh  hhgh hghgh

4:
hhhh hghhh hhghh hghghh hhhgh hghhgh hhghgh hghghgh

5:hhhhh hghhhh hhghhh hghghhh hhhghh hghhghh hhghghh hghghghh hhhhgh hghhhgh hhghhgh hghghhgh hhhghgh hghhghgh hhghghgh hghghghgh

n = 5 heads
Now the (1/2)2^n = 16 heads with/without gaps.
Then connect those with the similar 16 tails with/without gaps.
It's like zebra stripes of different widths.
I made a chart, multiplied up and added'm.

252 = 5 + 5 + 5 + 5 + 40 + 40 + 40 + 40 + 36 + 36 (my notes)

Last edited by John E. Franklin (2008-04-01 12:28:52)

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igloo myrtilles fourmis

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: How many bit strings of length 10:

(d) But it isn't 5! for the first choice.  If I have the binary string 1100111000, and I switch the first two bits, I get the string 1100111000, which is exactly the same.   You're counting duplicates.  So instead, it's 10 choose 5 (choose 5 bits which are 1's).  Now what about placing the 0's?

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: How many bit strings of length 10:

I tried to think of it like this, total subsets is 2^100 so if I want how many subsets with more than 2 elements then it would be:

2^100 - (subsets with no elements) - (subsets with 1 element) - (subsets with 2 elements)
2^100 - 2^0 - 2^1 - 2^2
2^100 - 1 - 2 - 4
2^100 - 7

You started off right, but you assumed there was a pattern when really none existed.

You have 100 objects to select from, and you want a collection of k objects.  How many ways are there to do this?  (Hint: Read my response to part (d) )

Hint #2: Do you really think there are only 2 ways to choose a subset of size 1 out of a set of 100 objects?  Intuitively, how many ways should there be?

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."