## PERMUTATIONS AND COMBINATIONSTopic 24, Section 2 - Combinations Factorial representation of combinations In permutations, the order is all important -- we count Here are all the combinations of
There are four such combinations. We call this The number of combinations of 4 distinct things taken 3 at a time. We will denote this number as
Now, how are the number of combinations Since the order does not matter in combinations, there are clearly fewer combinations than permutations. The combinations are contained among the permutations -- they are a "subset" of the permutations. Each of those four combinations, in fact, will give rise to 3 permutations:
Each column is the 3 permutations of that combination. But they are all
Example 1. How many combinations are there of 5 distinct things taken 4 at a time?
Again, both the numerator and denominator have the
Example 2. Evaluate
Both the numerator and denominator have 6 factors. The entire denominator cancels into the numerator. This will always be the case.
Example 3. Evaluate
We see that Always:
The bottom indices,
Example 4. Write out
The 3 factors in the numerator begin with Factorial representation In terms of factorials, the number of selections -- combinations -- of
Note also the convention that the factorial of the lower index,
Example 6. Write each of the following with factorials:
Therefore, we see that the number of ways of taking 6 things from 8, is the same as the number of ways of taking 2.
In general,
(See Problem 9, below.)
Example 7. Write
only 1 way to take 0 things from 8. This is the same as the number of ways of taking all 8.
Example 8. Write
The lower factorials are the Let us apply this to
Therefore,
a) Write all the combinations of To see the answer, pass your mouse over the colored area. a, b, c, d. b) Write their combinations taken 2 at a time. ab, ac, ad, bc, bd, cd. c) Write their combinations taken 3 at a time. abc, abd, acd, bcd. d) Write their combinations taken 4 at a time. abcd Problem 2. a) There are 3 permutations of the letters b) b) must the
b) only their * * *
Problem 3. You have 5 shirts, but you will take with you only 3 for your vacation. In how many different ways can you do this?
Problem 4. From a class of 12 students, 4 will be chosen to do a job. In how many different ways could that happen?
Problem 5. Evaluate the following.
Problem 6. a) Write out
b) In the numerator of
In part a) where c) In the numerator of * * *
Problem 7. Write the following with factorials.
Therefore, what number is Problem 8. Write the following with factorials.
a) In how many ways could you select three of these digits: 1, 2, 3, 4, 5 ?
b) In how many ways could you not select two of them?
c) Prove:
The sum of all combinations What is the sum of all possible combinations of We will see that the sum is equal to 2 For, consider any situation in which there are exactly two possibilities: Succeed or Fail. Yes or No. In or Out. Boy or Girl. And so on. Now suppose that we are going to make a choice between those two possibilities exactly Yes No Yes Yes. Another might be No No No Yes. Another might be Yes No Yes No. And so on. Let us ask: In how many of those outcomes will there be, for example, exactly 3 Yes's? To answer, consider that there will be 4 positions to fill with either Yes or No. In how many different ways, then, could we choose 3 positions to have a Yes? It will be the combinatorial number
Similarly, the number of outcomes with exactly 1 Yes will be The sum of those combinatorial numbers will account for the total number of ways to choose between two things 4 times.
What is that total number? Consider that to each choice of either Yes or No, there will be 2 possibilities, obviously. The first of the 4 choices can happen in 2 different ways. After that has happened, the second choice can also happen in 2 different ways, and similarly with the third and fourth choices. The total number of possible outcomes, then, is 2 The sum of all the combinations of 4 things is 2
There is exactly 1 way, Each combinatorial number That distribution of outcomes that consists of the sum of all possible combinations, is called a binomial distribution. It is called that because, as we will see in the next topic, the combinatorial numbers are the binomial coefficients. In general, the sum of all the combinations of
(Compare Problem 5, Topic 26.) Problem 10. a) If 5 children are born over the years, what is the total number of
2 b) How many of those sequences will have 0 boys, how many 1 boy, 1 5 10 10 5 1
These are the combinatorial numbers There will be 1 with no boys, 5 with 1 boy, 10 with 2, 10 with 3, 5 with 4, and 1 with 5. Problem 11. Of all possible outcomes on tossing a coin 6 times, how many of them will have heads 4 times?
Problem 12. At Joe's Pizza Parlor, in addition to cheese there are 8 different toppings. If you can order any number of those 8 toppings, then how many different toppings could you possibly order?
2 Problem 13. a) A door can be opened only with a security code that consists of five
There are b) If, to open the door you must press three codes consecutively, then
According to part a), there are 31 ways to choose the first code. Again, 31 ways to choose the second, and 31 ways to choose the third. Therefore, the total number of ways to open the door is 31 Next Topic: The binomial theorem Please make a donation to keep TheMathPage online. Copyright © 2015 Lawrence Spector Questions or comments? E-mail: themathpage@nyc.rr.com |