Permtations and Combinations: distributing gifts

[Louise Johnson]

In how many ways can 15 gifts be distributed equally:
a) amongst Kiana, Kaena, and Kalena

My answer \(\displaystyle {}_{15}C_3 = 455\)

b) into three parcels of five gifts each

My answer: \(\displaystyle {}_3C_5 = 10\)


They seem correct to me but I have probably missed something as per usual. This is also kinda on the same subject but is there an easy way to explain to me why
\(\displaystyle \L\\{}_nC_n = 1\) and
\(\displaystyle \L\\{}_nC_0 = 1\)
Thank you
Louise

 

[Trebor]

Firstly, If I know my stuff (and I really should), I believe that both a) and b) should be the same, since you are both distributing them into three groups equally, which implies that Kiana, Kaena and Kalena each recieve 5 gifts.

The order they recieve the gifts is not important, so you calculate the combinations not permutations for the answers.

Consider, the number of gifts you give the first group (person or parcel, the method is the same). You need to find the number of combinations of 5 gifts from 15. This will be the number of possible combinations of gifts in the first group. Then there will be 10 gifts left, and so for the second group you will to find, again, the number of combinations of 5 gifts from 10. For the last group, there will only be 5 gifts left, so there is only one possible combination.

Therefore, the answer is:

\(\displaystyle {}_{15}C_5 \times {}_{10}C_5 \times 1\)

Secondly, the reason why:

\(\displaystyle {}_nC_c = {}_nC_0 = 1\)

is that there is only one possible combination of n objects from a set of n, and that's all of them. Similarly, there is only one possible combination of 0 objects from a set of n, and that's none of them.

 

[pka]

Trebor is correct on part a. Although I would prefer my students to see the answer as \(\displaystyle \frac{{15!}}{{\left( {5!} \right)^3 }}\). That is the number of ways to arrange 5 X’s, 5 Y’s and 5 Z’s. The first girl gets the toys labeled X, second Y’s and third gets the Z’s. These are known as ordered partitions.

However, Trebor missed the second one. The answer is \(\displaystyle \frac{{15!}}{{\left( {5!} \right)^3 \left( {3!} \right)}}.\)
These are unordered partitions.

 

[soroban]

Hello, Louise!

The problems are getting trickier.
Don't feel bad if something subtle slips by you.
. . That's how we learn, you see,

In how many ways can 15 gifts be distributed equally:
. . a) amongst Kiana, Kaena, and Kalena

This is an "ordered partition".
There are: \(\displaystyle \L\:\begin{pmatrix}15\\5,5,5\end{pmatrix}\:=\;\frac{15!}{5!5!5!} \:=\:756,756\) ways

b) into three parcels of five gifts each

Unlike part (a), this is unordered partiaion.
The order of the parcels is not considered.

The answer is (a) must be divided by \(\displaystyle 3!\;\;\L\frac{756,756}{6}\:=\:126,126\)

Is there an easy way to explain to me why:
. . \(\displaystyle \L\\{}_nC_n = 1\) and
. . \(\displaystyle \L\\{}_nC_0 = 1\)

If there are \(\displaystyle n\) objects, in how many ways can we take \(\displaystyle n\) of them?
. . There is one way . . . take all of them.

If there are \(\displaystyle n\) objects, in how many ways can we take none of them?
. . There is one way . . . reject all of them.

 

[Louise Johnson]

Thank you to all of you for taking the time to post stuff. I am truely doing my best at learning and your posts are making a major difference! The questions are certainly getting tricky and I spend hours reading and studying posts and referring to them for reference!
Thank you
Sincerly
Louise
Ps Soroban that was an excellent explanation by the way and very easy to understand