MATH-372

Tucker Combinatorics Simplified

Chapter 5 – Enumeration

Section 5.2 – Example 1 –

There are two parts to this question :

Part (a) is to calculate how many ways there are to arrange 5 (distinct) objects.

and

Part (b) is to calculate the probability that a particular one will be in second place ?

Here's how (hopefully more intuitive than the text presents its solutions) :

a) How many ways are there to arrange n objects ?

First, use P(n, n) to arrange all of the objects, including the one specified, which is n! .

The books says that solving one problem may be of little help in solving another : :

“the solution to problem A is typically of

little help in solving problem B,”

… from book : Example 4 in Section 5.1

However, that is exactly what we WILL do in this case :

We will use the information from part (a) to help us solve part (b) .

b) What is the probability that a particular one will be in second place ?

To figure the probability that the specified object will be in second place, first subtract him from the group of n objects, which means there are now n-1 objects remaining.

Now, use P(n-1, n-1) to arrange these n-1 objects, which is how many ways that the specified object can be in second (or in any) place. This calculates how many places everyone else can be with it in second place, which is the same thing.

Now, use P(n-1, n-1) to arrange all of the objects except the specified object as the numerator, and also re‑use P(n, n) from part (a) of this problem, despite Tucker’s admonition (above), to arrange all of the objects, including the specified object, as the denominator. This forms a ratio of how many ways there are to arrange the objects without the specified object, P(n-1, n-1), divided by how many ways there are to arrange the objects with the specified object, P(n, n) :

P(n-1, n-1)

P(n, n)

This ratio calculates the probability that the specified object will be in second (or any) place.

Section 5.2 – Example 2 –

(a) How many ways are there to arrange the letters in the word “SYSTEMS” ?

The important point here is to remember to arrange the letters that can be re-arranged first.

Here, only 4 of the 7 letters are distinct in this regard.

The remaining 3 S’s are all the same, and therefore do not need arranging, right ?

Because there are 7 letters in all, first arrange the 4 distinct letters

into the 7 possible locations in the sequence.

Since you must use the letters given in the word “SYSTEMS,” there is no repetition, so that the first letter (whichever you choose) can go in any of 7 locations, the next letter goes into the remaining 6 locations,

then 5, then 4, for P(7, 4) = 7! / 3! which is (7 x 6 x 5 x 4 x 3 x 2 x 1) / (3 x 2 x 1) = 7 x 6 x 5 x 4 = 840 ways.

This leaves only 3 locations to put the 3 S’s, and there is only one way to do that,

so multiply 7 x 6 x 5 x 4 x 1 times 1.

So, the answer is 7 x 6 x 5 x 4 x 1, and the “1” doesn’t change anything, does it ?

(b) How many if the three S’s appear consecutively ?

This part is easier if you consider the consecutive S’s as a single character, “SSS.”

This means that there are really only 5 letters to arrange in the word, so P(5, 5) = 5!

Section 5.2 – Example 3 –

Here it is only necessary to notice that, since there are only two characters available, having n of one of them is the same as not having n of the other.

Therefore, the number of ways to arrange where the 0’s go is the same as the number of ways to arrange where the 1’s do NOT go.

Section 5.2 – Example 4 –

How many ways can 3 Aces be arranged in a hand.

First, consider that arranging 3 Aces is not the same as getting 3 Aces.

As usual, in probability questions, try not to think of the probability of getting 3 Aces, (which is .0181%), but simply how many ways (in this question) are there to arrange them, which is simply C(4, 3) = 4 ways. This assumes that you have them to arrange !

For example, the odds of me getting 5 Ferraris is much lower than the number of ways I could arrange them in driveway if I managed to get them, which is P(5, 5) = 5! / 0! = 5 x 4 x 3 x 2 x 1 / 1 (because q! = 1), or 120 ways.