# Math Help - please need help !!!!!!

1. ## please need help !!!!!!

Hello.. I have an assignment and I hope that you guys can help me a little bit if you dont mind. Basically all of the questions are related to counting. Any help would be appreciated. Even if you guys can just clarify or explain the question to me without solving that would be super. Thanks in advance.

Q1: In poker, a flush is a 5-card hand where all the cards have the same suit. Recall that a deck of cards has 52
cards. There are 4 suits (hearts, clubs, diamonds and spades) and each suit has 13 cards. How many different
flush hands can you have?

Q2:
How many 12 letter strings are there that contain at least 3 vowels (a vowel is one of
{a,e,i,o,u})?

Q3: How many bit strings of length 14 contain 7 consecutive 1’s or 7 consecutive 0’s?

2. Originally Posted by MMM88

Hello.. I have an assignment and I hope that you guys can help me a little bit if you dont mind. Basically all of the questions are related to counting. Any help would be appreciated. Even if you guys can just clarify or explain the question to me without solving that would be super. Thanks in advance.

Q1: In poker, a flush is a 5-card hand where all the cards have the same suit. Recall that a deck of cards has 52
cards. There are 4 suits (hearts, clubs, diamonds and spades) and each suit has 13 cards. How many different
flush hands can you have?

Q2:
How many 12 letter strings are there that contain at least 3 vowels (a vowel is one of {a,e,i,o,u})?

Q3: How many bit strings of length 14 contain 7 consecutive 1’s or 7 consecutive 0’s?

Q1. ${4 \choose 1}{13 \choose 5}= 5148$

Q3.

First we'll consider the string of 7 1's first.

Case 1
If the string of 1's is on the side:
[1111111] 1 * 2 * 2 * 2 * 2 * 2 * 2
2 * 2 * 2 * 2 * 2 * 2 * 1 [1111111]
The number immediately after or before can only be 0 (i.e. there is only one choice). For the rest of the numbers there are two choices (0 or 1).
The number of ways: $2(2^6)=128$

Case 2
If the string of 1's is NOT on the side, the numbers immediately to the left and right of it can only be 0's (i.e. there is only one choice). But the rest of the spaces can have 2 choices each.
e.g
2 * 1 * [1111111] * 1 * 2 * 2 * 2 * 2
Note that there are 6 ways the string can not be on the side, so
Number of ways: $6(2^5) = 192$

Without loss of generality, the string of 7 0's can be arranged the same number of ways as the string of 7 1's = 320 ways.

So in total we have 2(320) = 640.

But wait, we have counted some cases twice! The cases I'm referring to are:
[0000000][1111111]
[1111111][0000000]
We have counted each once for when we considered 7 1's in Case 1 and counted over them again when we considered 7 0's. So we need to subtract 2 from the total, giving us a final total of $\boxed{638}$.

3. Originally Posted by MMM88
Q2: How many 12 letter strings are there that contain at least 3 vowels (a vowel is one of {a,e,i,o,u}

Q3: How many bit strings of length 14 contain 7 consecutive 1’s or 7 consecutive 0’s?
There is an ambiguity in both problems.

In Q1 can any letter be used more than once? Is repartition allowed?
Could we use “bbbaaccccdde” as one of the strings?

Q2 is not clearly stated either.
Here are two 14-bit strings both containing 7 consecutive 1’s: 11111110000000 & 11111111111111. Do we count them both?

4. It makes a lot sense now... thanks alot guys.!!!