Counting Problems Math Help-3 word problems

Can someone help me with these questions? My math book IMO sucks and the wording is really poor at times.

1. How many ways can five different pairs of identical teddy bears be arranged in two rows of five for a photograph?

I tried solving this. However, I honestly don't even know what this is even asking.

I know you use permutations to solve, because order matters, right? I tried doing 10 P 5 X 5 P 5 but no luck. I assumed there were 10 bears in total, but IDK.

The answer is 113,400. I don't know how they got this.

2. Six different types of boats have pulled into a marina and want to dock at the six available slips. The six slips are adjacent to each other. How many ways can the six boats dock so that two cabin cruisers, which are travelling together, are docked next to each other?

The answer apparently is 240. I tried a lot of different solutions but nothings working.

3. How many different five-card hands that contain at most three hearts can be dealt from a standard deck of playing cards?

I think I am close to the solution here. I know what you have to do technically. I considered 4 separate cases:

0 hearts, 5 other suits

1 heart, 4 other suits,

2 hearts, 3 other suits,

3 hearts, 2 other suits

I know this is a combinations problem. I just don't know what my N values would be.

Like to solve, I thought:

13 C 0 X 39 C 5 + Case 2 + Case 3+Case 4...etc

but I am off by a lot. The answer is 2,569,788. Again IDK how they got this.

Please help!

Re: Counting Problems Math Help-3 word problems

Quote:

Originally Posted by

**MOMighty** 1. How many ways can five different pairs of identical teddy bears be arranged in two rows of five for a photograph?

The answer is 113,400. I don't know how they got this.

I absolutely agree with you about how poorly written these are.

But I can tell you how they got that answer.

Consider the string $\displaystyle AABBCCDDEE$.

That string can be arranged in $\displaystyle \frac{10!}{2^5}=113,400$ ways.

But each of those strings can be 'broken' into two rows of five each.

Re: Counting Problems Math Help-3 word problems

Quote:

Originally Posted by

**MOMighty** 3. How many different five-card hands that contain at most three hearts can be dealt from a standard deck of playing cards? The answer is 2,569,788. Again IDK how they got this.

$\displaystyle \sum\limits_{k = 0}^3 {\binom{13}{k}\cdot\binom{39}{5-k}} = {{2569788}}$

Re: Counting Problems Math Help-3 word problems

You see, I've never been taught that before in this book.

The repetition formula I learned is is n!/a!b!c!...etc.

where n!=the total number of objects

and a!b!c!=the number of objects that repeat

EDIT: Wait nevermind!

I get it in context of what I learned.

10!/(2!x2!x2!x2!x2!)=113400

Is there anyway to relate that question to the above formula? I don't recognize your solution.

Oh and for question two the answer is 240, my bad!

Thanks.

Re: Counting Problems Math Help-3 word problems

Quote:

Originally Posted by

**MOMighty** You see, I've never been taught that before in this book.

The repetition formula I learned is is n!/a!b!c!...etc.

where n!=the total number of objects

and a!b!c!=the number of objects that repeat

Is there anyway to relate that question to the above formula? I don't recognize your solution.

It is the same formula. Note that $\displaystyle 2!\cdot2!\cdot2!\cdot2!\cdot2!=2^5$

Re: Counting Problems Math Help-3 word problems

Yeah, I just got that haha.

For what it's worth, another person told me the solutions he came up with for questions 2 and 3. His solution for 3 is wrong according to the book, but for two he said:

2.) Consider at first the two cabin cruisers as one entity. Then we have 5 objects

to order, which can be done in 5! ways. Now for each of these orderings the

two cabin cruisers can be arranged in two ways, so the final answer is 2*5! = 240.

Another way to look at it is that there are 5 'pairs' of adjacent slips for

the two cabin cruisers that are traveling together to dock at. For each of these

choices, the remaining 4 boats can be arranged in 4! = 24 ways, and for each

of these arrangements the two cabin cruisers can be arranged in two ways.

This give a total of 5 * 24 * 2 = 240 possible arrangements. This is not, however,

the answer you have been provided with.

**The thing I don't understand about this solution is why the cabin cruisers can only be arranged in two ways. That makes no sense to me.**

I thought it was five ways.

6 different slips.

CCBBBB (one way)

BCCBBB (two ways)

BBCCBB (three ways)

BBBCCB (four ways)

BBBBCC (five ways)

with B=normal boat

and C=cabin cruiser

3.) The easiest way to do this is to first find the number of 5-card hands that

have 4 hearts. This will be (13 C 4)*(39 C 1), since we need to choose 4 of the

13 hearts and then 1 of the 39 remaining cards which are not hearts. Now subtract

this from the total number of possible 5-card hands to get the number of 5-card

hands with at most 3 hearts. So the final answer is

(52 C 5) - (13 C 4)*(39 C 1) = 2598960 - 715*39 = 2571075 hands.

**I recognize this solution but I am positive there are two solutions to solve this problem and my solution is one of the ways.**

Honestly, this book makes me so frustrated...:(

It was just this chapter that gave me trouble. IDK.

Re: Counting Problems Math Help-3 word problems

Sorry for the double post, but I got the solution for question 3.

My solution was indeed correct. I just screwed up the math.

Four separate cases.

0 hearts, 5 other suits

1 heart, 4 other suits

2 hearts, 3 other suits

3 hearts, 2 other suits

(13 C 0)(39 C 5) + (13 C 1)(39 C 4) + (13 C 2)(39 C 3) + (13 C 3)(39 C 2)

=2,569,788

Re: Counting Problems Math Help-3 word problems

Hello, MOMighty!

Quote:

2. Six different types of boats have pulled into a marina and want to dock at the six available slips. .The six slips are adjacent to each other. .How many ways can the six boats dock so that two cabin cruisers, which are travelling together, are docked next to each other?

. . The answer is 240.

Call the boats: $\displaystyle A,B,C,D,E,F.$

Suppose $\displaystyle A$ and $\displaystyle B$ must adjacent.

Duct-tape them together: $\displaystyle \boxed{AB}$

Then we have five "boats" to arrange: .$\displaystyle \boxed{AB}\;C\;D\;E\;F$

There are $\displaystyle 5! = 120$ ways.

But $\displaystyle \boxed{AB}$ could also be $\displaystyle \boxed{BA}.$

Therefore, there are: .$\displaystyle 2 \times 120 \:=\:240$ ways.

Quote:

3. How many different 5-card hands with at most 3 Hearts can be dealt from a standard deck of playing cards?

. . The answer is 2,569,788.

There are: .$\displaystyle _{52}C_5 \,=\,2,\!598,\!960$ possible hands.

There are 13 Hearts and 39 Others.

The opposite of "3 or less" is "4 or more."

How many hands have 4 Hearts or 5 Hearts?

4 Hearts, 1 Other: .$\displaystyle (_{13}C_4)(_{39}C_1) \:=\:\frac{13!}{4!\,9!}\cdot\frac{39!}{1!\,38!} \:=\: 27,\!885$

5 Hearts, 0 Others: .$\displaystyle (_{13}C_5)(_{39}C_0) \:=\:\frac{13!}{5!\,8!}\cdot1 \;=\;1,\!287$

Hence, there are: .$\displaystyle 27,\!885 + 1,\!287 \:=\:29,\!172$ hands with 4 or 5 Hearts.

Therefore, there are: .$\displaystyle 2,\!598,\!960 - 29,\!172 \:=\:2,\!569,\!788$ hands with at most 3 Hearts.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Your game plan was *excellent* . . . It should have worked.

$\displaystyle \begin{array}{ccccc}\text{0 H, 5 Others:} & (_{13}C_0)(_{39}C_5) &=& \;\;\;575,757 \\ \text{1 H, 4 Others:} & (_{13}C_1)(_{39}C_4) &=& 1,069,263 \\ \text{2 H, 3 Others:} & (_{13}C_2)(_{39}C_3) &=& \;\;\;712,842 \\ \text{3 H, 2 Others:} & (_{13}C_3)(_{39}C_2) &=& \;\;\;211,926 \\ \hline & \text{Total:} && 2,569,788 \end{array}$

Re: Counting Problems Math Help-3 word problems

Thank you sir!

Do you think it is valid if I see the question like this instead?

5 separate cases, like I thought out:

CCBBBB (one way)

BCCBBB (two ways)

BBCCBB (three ways)

BBBCCB (four ways)

BBBBCC (five ways)

with B=normal boat

and C=cabin cruiser

In one case, it is always

2 X 1 X 4 x 3 x 2 x 1=48

48 x 5 separate cases=240 ways

That's the only way I can think of when taking into account what I've been taught. I had just read the question wrong I believe.

What do you think?

Re: Counting Problems Math Help-3 word problems

How can I learn math word problems to teach my child for homework?

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