Room assignments

**oldguynewstudent** · July 31st 2010, 07:12 AM

Last problem of the semester! Could someone let me know if the following is correct?
Thanks

A university has 120 incoming freshman that still have to be assigned to on-campus housing. The only remaining dorm holds 105 students and contains 42 doubles and seven triples. In how many ways can the university select 105 students to house in this dorm and then arrange those students into roommate pairs and triples, without yet assigning them to rooms?

$\left({120\atop 105}\right)*\left({105\atop 2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2, 2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3}\r ight)$

**Soroban** · July 31st 2010, 07:42 AM

Hello, oldguynewstudent!

I agree with your reasoning and your answer.

A university has 120 incoming freshman that will be assigned to on-campus housing.
The only remaining dorm holds 105 students and contains 42 doubles and 7 triples.
In how many ways can the university select 105 students to house in this dorm
and then arrange those students into roommate pairs and triples?

$\left({120\atop 105}\right)\!\cdot\!\left({105\atop 2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2, 2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3}\r ight)$

First, choose 105 from the 120 incoming freshmen.

. . $\dfrac{120!}{105!\,15!}}$ ways.

Second, partition the 105 students into 42 pairs and 7 triples.

. . $\dfrac{105!}{(2!)^{42}(3!)^7}$ ways.

Finally, the answer is the product of the two quantities above.

I'll let you crank out the answer . . . I'll wait in the car.

**Plato** · July 31st 2010, 07:50 AM

Soroban, did you notice the phrase "without yet assigning them to rooms?"
It seems to me to mean unordered partitions.
The answer you gave is for ordered partitions.
That is assigning them to rooms.

I do not understand oldguynewstudent's notation.

**oldguynewstudent** · July 31st 2010, 08:13 AM

Originally Posted by Plato

Soroban, did you notice the phrase "without yet assigning them to rooms?"
It seems to me to mean unordered partitions.
The answer you gave is for ordered partitions.
That is assigning them to rooms.

I do not understand oldguynewstudent's notation.

I am thinking of this as distributing identical housing to distinct students. The identical housing comes in groups but we haven't specified which group is which yet.

The notation is a multinomial notation, like binomial but with multiple groups. Does that make sense? BTW, I want to thank you for all the wonderful help you have given me so far. (Gesturing "I'm not worthy!")

**Plato** · July 31st 2010, 08:22 AM

How does one calculate $\binom{10}{2,3.5}$ ?

**oldguynewstudent** · July 31st 2010, 08:37 AM

Originally Posted by Plato

How does one calculate $\binom{10}{2,3.5}$ ?

10!/(2!3!5!)

**oldguynewstudent** · July 31st 2010, 08:45 AM

For any $n\geq0$ ,

$\left(x_{1}+x_{2}+...+x_{k}\right)^{n}$ $=\sum\left({n\atop t_{1},t_{2},...,t_{k}}\right)x_{1}^{t_{1}}x_{2}^{t _{2}}...x_{k}^{t_{k}}$ where the sum is over all k-lists $\left(t_{1},t_{2},...,t_{k}\right)$ of nonnegative integers that sum to n.

**Plato** · July 31st 2010, 09:05 AM

So that is the number of ways to divide 10 students into a group of 2, a group of 3 and a group of 5.
Well how many of ways to divide 18 students into three groups of 2 and two groups of 6?

**Soroban** · July 31st 2010, 09:36 AM

Originally Posted by Plato

Soroban, did you notice the phrase "without yet assigning them to rooms?"
It seems to me to mean unordered partitions.
The answer you gave is for ordered partitions.
That is assigning them to rooms.

You're right, Plato! . . .I missed the significance of that phrase.

My number: . $\dfrac{105!}{(2!)^{42}(3!)^7}$ .is the number of ordered partitions.

To un-order them, we would divide by $(42!)(7!)$

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