Perms and Combs (1 Viewer)

[pomsky]

The ratio of the number of combinations of (2n+2) different objects taken n at a time to the number of combinations of (2n-2) different objects taken n at a time is 99:7. Find the value of n.

Ans: N= 5
How do I approach these type of questions? And when do I use factorial, perms or combs? I'm getting them all mixed up :s

 

[VBN2470]

Essentially you want to solve

.

Use the definition of to find your value of .

 

[pomsky]

Essentially you want to solve

.

Use the definition of to find your value of .

Ohh- I geddit it now
Thank you!

 

[pomsky]

But why would it be P and not C?

 

[VBN2470]

Sorry, it should be C, I misread the question. In combinations, the order in which the elements are arranged is disregarded, unlike in permutations where the order matters.

 

[pomsky]

Thanks VBN
How does factorial fit into the mix?

 

[VBN2470]

Thanks VBN
How does factorial fit into the mix?

Factorials are used when considering order of objects so if I have a set {A, B, C} then the total number arrangements are 3 X 2 X 1 = 3! = 6. Here, repetition is not allowed. With combinations, you divide the total number of permutations by the total number of arrangements for the elements being considered. For example, there are 3C2 ways to pick a group of two elements from the set namely {A, B}, {A, C} and {B, C}, BUT the total number of ordered sets that exist is 3C2 X 2! = 6. These ordered arrangements are then {A, B}, {B, A}, {A, C}, {C, A}, {B, C} and {C, B}, where the 2! is used to arrange the elements in each subset.

 

[VBN2470]

NOTE that nPr and nCr are just definitions that you need to follow, with the relation that nPr = r! X nCr, since you arranging each of the r elements in your chosen subset to obtain all ORDERED sets with the same number of r elements in them. Elements can be different i.e. {A, B}, {A, C}, {B, C} each have different elements in them, but they all have 2 (= r) elements in each subset and when you arrange them, you will get {A, B}, {B, A}, {A, C}, {C, A}, {B, C} & {C, B}.

 

[VBN2470]

Don't know if my explanation is clear cut, you will need read the textbook so you can follow more detailed explanations as to knowing when to use what. Cambridge seems to explain these things quite well, so have a look at that if you still don't understand

 

[pomsky]

Thanks again VBN

Am I using the principles right in the following Q:
How many different selections of four letters may be made from the word WEDNESDAYS and in how many ways can they then be arranged in a row?



Ans: 83 which is a far cry from the 640 I'm getting @@

 

[VBN2470]

So it should be 7C4 + (3C1 X 6C2) + 3C2 = 83? This is just referring to the different selections (combinations) only.

 

[pomsky]

So it should be 7C4 + (3C1 X 6C2) + 3C2 = 83? This is just referring to the different selections (combinations) only.

Could you explain how you got to that?
Is it by taking cases: no vowels, 1 vowel, 2 vowel?

 

[VBN2470]

Could you explain how you got to that?
Is it by taking cases: no vowels, 1 vowel, 2 vowel?

Not about vowels, but about taking repetitions of letters. We have WEDNESDAYS, so we can say that the total number of distinct letters is just W, E, D, N, S, A, Y. There are 7 elements here, so if we want to pick 4, there will be 7C4 total possibilities (this ensures no letter will be repeated in any chosen set). Now the repeated letters are D, E & S, so if we want to pick a set of 4 which exactly one set of repeated letters, this can be done in 3C1 ways (selecting one of the three letters can be repeated twice) X 6C2 (selecting the remaining letters). NOTE that the 6 comes from the fact that we already picked one letter for repetition, so instead of having 7 distinct letters now, we have 6. For two sets of repeated letters, e.g. D D E E or D E D E or E S S E (combination, where order doesn't matter) there are 3 such letters that can be repeated twice (D, E, S) and we want two, so there are 3C2 = 3 ways of picking this combination. Then we add all the different cases to get the total number of ways in which we have all possible selections for picking ANY 4 elements from the given set.

I hope that helped

 

[VBN2470]

Vowels will come into play if the question actually specified something about it, in this case it is just general for any set of 4 letters you pick from the original set.