## PERMUTATIONS AND COMBINATIONSTopic 24, Section 2 - Combinations Factorial representation of combinations If we have 10 letters 10 Say, however, that we do not care about the
—are the same combination. How many combinations, then, are there of 10 things taken 4 at a time? For each combination, there are 4! permutations. In other words, the numbers of permutations is 4! times the number of combinations. Therefore, the number of combinations is equal to the number of permutations divided by 4!.
This is The number of combinations of 10 distinct things taken 4 at a time. We will denote it by the symbol
The upper index Example 1. You have 6 pairs of socks, but on your vaction you will take only 3. In how many different ways could you do that?
Again, both the numerator and denominator have the
Example 2. Evaluate
The numerator and denominator will always have common factors. The number of combinations is always a whole number.
Example 3. Evaluate
We see that Always:
The bottom indices,
Example 4. Evaluate
Example 5. Write out
The 3 factors in the numerator begin with Factorial representation In terms of factorials, the number of combinations of
This is
Note also the convention that the factorial of the lower index,
Again: 5 + 3 in the denominator equals 8 in the numerator.
Example 7. Write each of the following with factorials:
But 2 6 is equal to 6 2. We see again that the number of ways of selecting 6 things from 8, is equal to the number of ways of selecting 2.
(See Problem 8, below.)
Example 8. Write
Since 0 = 1, that fraction is equal to 1. There is only 1 way to take 0 things from 8. This is the same as the number of ways of taking all 8.
Example 9. 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.
* * *
Problem 3. From a class of 12 students, 4 will be chosen to do a job. In how many different ways could that happen?
Problem 4. Evaluate the following.
Problem 5. a) Write out
b) In the numerator of
In part a) where c) In the numerator of * * *
Problem 6. Write the following with factorials.
Therefore, what number is Problem 7. 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 it is equal to 2 Let us consider a specific case, say all the possible combinations of 5 distinct things: There will, for example, be combinations taking 2 of them:
There will be combinations taking 3:
There will be one combination with all 5:
In each combination, a letter will be included or it will not. In the combination We can therefore construct a combination as follows. For each of the 5 letters, we will include it or not. For 2 They include the combination in which there are no letters, which is because we have said "no" to each one, and they move on to the combination in which we have said "yes" to every one. That is,
The actual values are as follows: 1 + 5 + 10 + 10 + 5 + 1 (See Pascal's triangle, Topic 24) There is exactly 1 way, 5 ways in which there is 1; 10 ways in which there are 2; 10 ways in which there are 3; 5 ways in which there are 4; 1 way in which there are 5. As we have seen, the lower index The sum of those combinatorial numbers is 2 The sum of all possible combinations of
(Compare Problem 5, Topic 26.) A binomial distribution We will now consider any case in which there are exactly two possibilities: Heads or Tails. Yes or No. Boy or Girl. And so on. We will see that that leads to what is called a binomial distribution. What will be the outcomes when one of those two possibilities happens One outcome of 4 tosses might be H T H H Another might be T T T H Another: H T H T And so on. Let us ask: In how many of those outcomes will there be, for example, exactly 3 H's? To answer, consider that there will be 4 positions to fill with either H or T. In how many different ways, then, could we choose 3 positions to have an H? It will be the combinatorial number
Similarly, the number of outcomes with exactly 1 H will be The lower index
That number is 2 The value of each of those combinatorial numbers is as follows: 1 + 4 + 6 + 4 + 1 There is exactly 1 way, 4 ways in which there is 1 H; 6 in which there are 2; 4 in which there are 3; 1 in which there are 4. (In the theory of probability, the list of all possible outcomes is called the sample space.) When there is one of two possibilities, the distribution of outcomes—the number of successes in Problem 9. Of all possible outcomes on tossing a coin 6 times, how many of them will have tails 4 times?
Problem 10.
2
1 5 10 10 5 1 There will be 1 with no boys, 5 with 1 boy, 10 with 2, 10 with 3, 5 with 4, and 1 with 5.
Those are the combinatorial numbers See Topic 25. Problem 11. 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 12.
There are
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 © 2021 Lawrence Spector Questions or comments? E-mail: teacher@themathpage.com |