Permutation and Combination Homework Help

[flaming1]

Hey guys, there's these questions in my hm that's really tough for me and i basically have no clue on how to do it. Any help is greatly appreciated
1.a) Give the numbers 1,1,2,3,4,5,5,6,7, If five numbers are randomly selected without replacement, how many different numbers can be formed if identical numbers can be selected? How many numbers in a) are even?

2. 5 boy and 5 girls were to sit in a row. No boy may sit next to a boy and no girl may sit next to a girl. How many ways are there to arrange the seats if Jennifer must be sitting next to jacky?

3. A bridge hand consists of 13 cards. How many bridge hands include 5 cards of one suit, 6 cards of a second and 2 cards of a third?

Thanks for the help :)

 

[Jameson]

Hi flaming, (Wave)

Welcome to MHB!

We ask that you post one problem per thread and that you show us what you have tried first. Let's take a look at #1.

So pretend that these are all different numbers. How many 5 digit numbers could we make? If you start with that then we can address the doubles.

 

[soroban]

Hello, flaming!

3. A bridge hand consists of 13 cards. .How many bridge hands include
. .5 cards of one suit, 6 cards of a second and 2 cards of a third?

There are $4$ choices of suit for the set-of-5.
There are: $$\:{13\choose5}$$ ways to choose the set.

There are $3$ choices of suit for the set-of-6.
There are: $$\:{13\choose6}$$ ways to choose the set.

There are $2$ choices of suit for the set-of-2.
There are: $$\;{13\choose2}$$ ways to choose the set.There are: $\:4{13\choose5}\cdot 3{13\choose6}\cdot2{13\choose2}$

$\qquad \:=\:4,\!134,\!297,\!024$ such bridge hands.

 

[soroban]

Hello again, flaming!

1. Given the digits 1,1,2,3,4,5,5,6,7.
If five digits are randomly selected without replacement,
how many different numbers can be formed?

There are 7 distinct digits; two of them are duplicated.

There are 3 cases to be considered.[1] Five distinct digits: ABCDE
$\qquad _7P_5 \,=\,2520$ ways.[2] One pair: AABCD
$\qquad$2 choice for the pair
$\qquad$Select 3 more digits from the other 6: ${6\choose3}$
$\qquad$Permute the 5 digits: ${5\choose2}$
$\qquad 2{6\choose2}{5\choose2} \,=\,400$ ways.[3] Two pairs: AABBC
$\qquad$We have {1,1} and {5,5}.
$\qquad$and 5 choices for the last digit.
$\qquad$Permute the 5 digits: ${5\choose2,2}$
$\qquad 5{5\choose2,2}\,=150$ waysAnswer: $2520 + 400 + 150 \;=\;3070$

 

[Jameson]

I did this a little differently and am trying to see which method is correct. My solution, if correct, is much simpler but I'm thinking it over.

We start with \(\displaystyle _9P_5\) and need account for all the double counting of the 1's and 5's so we divide by 2! two times. I get the following answer:

\(\displaystyle \frac{_9P_5}{2!2!}=\frac{15120}{4}=3780\)

If anyone has some insight into this I would be very curious to read it.

EDIT: I see Opalg might be responding to this question but just wanted to add that now I'm in doubt of my solution because we aren't using all the numbers all the time. If we were asked to make a 9 digit number then I am sure this method would work but there will be numbers that don't have a 1 or a 5 in them and dividing these by 2 as well seems incorrect.

 

[Opalg]

Sorry to say that I disagree with both those answers. I think that Jameson's method does not work for the reason he suggests in the Edit. As far as I can see, soroban's method is correct but there seems to be a numerical error.

[2] One pair: AABCD
$\qquad$2 choice for the pair
$\qquad$Select 3 more digits from the other 6: ${6\choose3}$
$\qquad$Permute the 5 digits: ${5\choose2}$
$\qquad 2{6\choose2}{5\choose2} \,=\,400$ ways.

In the last line, $6\choose2$ is clearly a misprint for $6\choose3$, but that is immediately corrected in calculating the number $400$. However, the next item "$5\choose2$" ought to be counting permutations, not combinations, and should therefore be $\dfrac{5!}2 = 60$, making the total for case [2] $2400$ instead of $400$. So I get the overall result to be $2000$ greater than soroban's, namely $5070$.

Anyone want to come up with yet another answer?