# Combinations of n objects taken r at a time

• Justin12005
In summary, the customer has 8 choices for the base and top, and 5 choices for each of the 4 modules in the middle. Therefore, the total number of choices for the completed file is 8 x 5 x 5 x 5 x 5, which equals 8,000 possible combinations.
Justin12005
An office furniture manufacturer that makes modular storage files offer its customers two choices for the base and four choices for the top, and the modular storage files come in five different heights. The customer may choose any combination of the five different-sized modules so that the finished file has a base, a top, and one, two, three, four, five, or six storage modules. How many choices does the customer have if the completed file has four storage modules, a top, and base? The order in which the four modules are stacked is irrelevant.

The best i could get was nCr=(n/r), (2/1)x(4/1)x((5/4)+(5/3)+(5/2)+(5/1)), and it is wrong. my teacher spent half an hour on this one question and he has no clue how to do it. This is basically my last resort.

It's like the ice-cream flavors problem - only the number of choices for top and bottom flavor are not the same as for the middle ones. You've done that one before right?
It would help to see your reasoning - but it looks like you did something to do with combinations out of 5 (but it looks like you got the formula mixed up) , times the 2 bases, times the 4 tops? The answer cannot be correct because it gives you a fraction in the answer - and you cannot have a fraction of a combination. In fact, the method you tried will usually give you a fraction. (BTW: do you know the correct answer?)

In the description:
You have a base (out of 2), a top (out of 4) so that's 8 right there... well done.
The middle part has 4 modules in the middle, which the customer assembles out of 5 varieties.
If the varieties are A,B,C,D,E, then one choice may be AABE, another may be ACCE.
If order matters, then AABE is different from ABAE etc.

Is the way forward clearer now?

Note: it is easy to get mixed up about formulas, so try to look at what the formulas are trying to describe:
http://www.mathsisfun.com/combinatorics/combinations-permutations.html

Last edited:

## 1. What is the formula for calculating combinations of n objects taken r at a time?

The formula for calculating combinations is nCr = n! / (r!(n-r)!), where n represents the total number of objects and r represents the number of objects taken at a time.

## 2. How is a combination different from a permutation?

A combination is a selection of objects where the order does not matter, while a permutation is a selection where the order does matter. In other words, combinations are unordered, while permutations are ordered.

## 3. Can the number of combinations be larger than the number of objects?

No, the number of combinations cannot be larger than the number of objects. The maximum number of combinations is nCr, where n is the total number of objects and r is the number of objects taken at a time.

## 4. How does the value of r affect the number of combinations?

The value of r directly affects the number of combinations. As r increases, the number of combinations decreases. This is because as r increases, the number of objects taken at a time decreases, resulting in fewer possible combinations.

## 5. How can combinations be used in real-life scenarios?

Combinations are commonly used in real-life scenarios such as lottery numbers, combinations for locks, and selecting a team from a larger group of people. They can also be used in statistics and probability to determine the likelihood of certain outcomes.

Replies
3
Views
881
Replies
10
Views
3K
Replies
14
Views
1K
Replies
3
Views
2K
Replies
28
Views
5K
Replies
3
Views
1K
Replies
3
Views
1K
Replies
1
Views
2K
Replies
1
Views
866
Replies
3
Views
3K