permutations with some identical items

xo_caroline_xo

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question: Ten students have been nominated for the positions of secretary, treasurer, social convenor, and fundraising chair. In how many ways can these positions be filled if the Norman twins are running and plan to switch positions on occasion for fun since no one can tell them apart?

so far i'm using the formula;

n!
a!b!c!

however, i dont know what numbers i should plug for the variables, because when i did substitute n = 10 (since there are 10 students) over (divide) by 2 since there are 1 sets of twins, i got the answer 1 814 400 but, the answer of the back of the book is 6144...any advice would be appreciated...thanks!

caroline
 
They had to throw twins into the mix... :shock:

If you don't include either of the twins, how many orderings can you get? (I think this is "<sub>8</sub>P<sub>4</sub>".)

If you regard the twins as one person (that is, if only one of the twins is picked, and the two of them swap out occasionally, but nobody knows, so it doesn't really "count"), then how many permutations can you get, each of which includes the twin? (I think you'd use "4<sub>8</sub>P<sub>3</sub>" for this, since you'd be picking three of the positions from the eight non-twins, and there are four choices for where the twin could go.)

Then how would it work if both twins were used? Since they'd be switching between two positions, and since they'd be indistiguishable, I think you'd use "<sub>8</sub>P<sub>3</sub>" to chose the other three people. Then you'd multiply by the number of ways to permute five objects, with three the same (the non-twins) and two the same (the twins).

But this gets me a total of 6384, not 6144. So the above may be "in the ballpark", and might be a good start, but isn't quite correct.

Anyway, I hope you can build a bit on this.

Eliz.
 
Hello, xo_caroline_xo!

I'd love to see the book's explanation for this problem (if there is one).
I don't understand where or how they got 6144 ... the answer is much smaller.

And if the Norman twins are totally indistinguishable, that makes the answer even smaller!


Ten students have been nominated for the positions of secretary, treasurer,
social convenor, and fundraising chair.
In how many ways can these positions be filled if the Norman twins are running
and plan to switch positions on occasion for fun since no one can tell them apart?

Forget the Norman twins . . .
Suppose there are ten different students nominated.

To fill the position of Secretary, there are 10 possible choices.

To fill the position of Treasurer, there are 9 possible choices.

To fill the position of Social Convener, there are 8 possible choices.

To fill the position of Fundraising Chair, there are 7 possible choices.


Therefore, there are: \(\displaystyle 10\,\times\,9\,\times\,8\,\times\,7\:=\:\)5,040 ways to fill the positions . . . maximum!

 
soroban said:
Suppose there are ten different students nominated....

Therefore, there are: \(\displaystyle 10\,\times\,9\,\times\,8\,\times\,7\:=\: 5,040\) ways to fill the positions . . . maximum!
I would agree -- if the twins weren't involved. If you had ten distinguishable people and were choosing four, then <sub>10</sub>P<sub>4</sub> = 5040 would work. But with the twins involved, you don't exactly have nine (distinguishable) people; you have ten people (by naming), nine of whom happen to be distinguishable (by appearance).

I dunno... I'm really bad at this sort of thing... But maybe it means something, that my total (6384) differs from the book's answer (6244) by 240, which is <sub>10</sub>C<sub>4</sub>...?

Eliz.
 
soroban said:
Hello, xo_caroline_xo!

I'd love to see the book's explanation for this problem (if there is one).
I don't understand where or how they got 6144 ... the answer is much smaller.

And if the Norman twins are totally indistinguishable, that makes the answer even smaller!

one down fall in my math textbook is that they dont show step by step methods in finishing a question, they just give you the final answer :(
 
stapel said:
I'm really bad at this sort of thing... But maybe it means something, that my total (6384) differs from the book's answer (6244) by 240, which is <sub>10</sub>C<sub>4</sub>...?

my teacher hasnt taught us how to use the nCr key on our calculator, is it still alright to use it tho? :?
 
xo_caroline_xo said:
[M]y teacher hasn[']t taught us how to use the nCr key on our calculator[. I]s it still alright to use it...?
I can't think why not. But some educators do have some pretty weird policies....

Eliz.
 
im still not getting the answer...this is quite depressing :cry: \

anyone else who can help me...please DO!!
 
xo_caroline_xo said:
im still not getting the answer...this is quite depressing
anyone else who can help me...please DO!!
You have been told that the book's answer cannot be correct.
If we have 10 unique people the maximum number is \(\displaystyle \frac{{10!}}{{6!}} = 5050.\)

The correct answer to your problem is \(\displaystyle \frac{{9!}}{{5!}} + \frac{{\left( {4!} \right)\left( {8!} \right)}}{{\left( {2!} \right)^2 \left( {6!} \right)}} = 3360.\)

If you need more exclamation, we can help.
 
If we have 10 unique people the maximum number is \(\displaystyle \left( {10} \right)\left( 9 \right)\left( 8 \right)\left( 7 \right) = \frac{{10!}}{{6!}}\)

You see we have 10 choices for president, 9 choices for vice-president, 8 choices for secretary, and 7 choices for treasurer.
 
the answer is incorrect since the answer is 6144 according to my teacher the answer in the back of the book is correct...what are we doing wrong?! :x
 
I, too, arrive at 3360 by counting the number of ways to choose the 4 out

of 10, assuming order matters.

\(\displaystyle P(10,4)=5040\)

Subtract the case where the Norman twins are not chosen. \(\displaystyle P(8,4)=1680\)

5040-1680=3360

I would also like to see your teacher's explanation, as well as the books interpretation of this problem.

The others are saying that there is a total of 5040 ways to choose 4 ordered objects from a group of 10 objects. It stands to reason, if the Norman twins are indistinguishable, that this number will be less than 5040. You then have 8 distinguishable and 2 indistinguishable objects. 6144 is higher still than 5040. Quite illogical. :roll:
 
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