# Thread: help me with a few counting problems

1. ## help me with a few counting problems

1) how many bit string of length 12 contain at least three 1s?
2) A coin is flipped 10 times where each flip comes up either heads or tails. how many possible outcomes contain at most three tails?

3) thirteen people on a softball team show up for a game. How many ways are there to choose 10 players to take the field?

4) a club has 25 members. How many ways are there to choose four members of the club to serve on an executive committee?

5) how many ways are there for a horse race with four horses to finish if ties are possible? (any number of the four horses may tie).
The answer to 3) should be:
C(10,9) + C(3,1)
OR
C(10,9)C(3,1) + C(10,8)C(3,2) + C(10,7) C(3,3)

The answer to 4) Is it permutation or combination? I thought order does matter in this case because no member would repeat for the same position, right?

For 5) and I have 4! for 1st place, second place, third place and fourth place.
1st position tie : 4 ways (as one to all of them could tie)
2nd position tie: 3 ways
3rd position tie: 2 ways
therefore, 24 + 9 = 23 ways? is it correct? if not, please explain me why?

2. Here is #5.
I will remind you that $\text{Surj}(n,k) = \sum\limits_{j = 0}^k {\left( { - 1} \right)^j \binom{k}{j}\left( {k - j} \right)^n }$
is the count of the number of surjections from a set of n to a set of k.

If I understand your horse race correctly the answer is $\sum\limits_{j = 1}^4 {\text{Surj}(4,j)}$.
The reason being: if all tie then all are in first place so $n=4~\&~k=1$
We can have a first place and a second place if three tie or two pairs tie: $n=4~\&~k=2$
Etc.

3. it does not work with either permutation or combination?

4. Originally Posted by zpwnchen
Consider all the 12-bit strings. There are $2^{12}$ of them.
How many of those have no 1's?
How many of those have exactly one 1's?
How many of those have exactly two 1's?

Now you must do some of this for yourself.

5. Originally Posted by Plato
Consider all the 12-bit strings. There are $2^{12}$ of them.
How many of those have no 1's? 1?
How many of those have exactly one 1's? C(12,1) = 12 ?
How many of those have exactly two 1's? c(12,2) = 66 ?

Now you must do some of this for yourself.
I gave my best shot!

for #5,
Cant it be done with either permutation or combination?