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12.2 Combinations and Binomial Theorem. p. 708 . In the last section we learned counting problems where order was important.
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In the last section we learned counting problems where order was important • For other counting problems where order is NOT important like cards, (the order you’re dealt is not important, after you get them, reordering them doesn’t change your hand) • These unordered groupings are called Combinations(objects arrangements are order unrelated, only the selected content is important)
A Combination is A selection of r objects from a group of n distinct objects where order is not important
Combination of n objects taken r at a time • The number of combinations of r objects taken from a group of n distinct objects is denoted by nCr and is:
nCr can be read “r combinations taken from a group of n objects” or “___________________”
If there are 5 different colored marbles in a bag, how many ways can you pick a group of 2 of them? =
Finding Combinations • In a standard deck of 52 cards there are 4 suits with 13 of each suit. • If the order isn’t important how many different 5-card hands are possible? • The number of ways to draw 5 cards from 52 is = 2,598,960
In how many of these hands are all 5 cards the same suit? • You need to choose 1 of the 4 suits and then 5 of the 13 cards in the suit. • The number of possible hands are:
How many 7 card hands are possible? • How many of these hands have all 7 cards the same suit?
Examples using Combinations In how many ways can you select 5 cards that are all Diamonds? In how many ways can you select 2 cards that are the same suite? In how many ways can you select 10 cards that are all the same suite?
When finding the number of ways both an event A _____ an event B can occur, you _____. • When finding the number of ways that an event A ___B can occur, you _____
Deciding to + or * • A restaurant serves omelets. They offer 6 vegetarian ingredients and 4 meat ingredients. • You want exactly 2 veg. ingredients and 1 meat. How many kinds of omelets can you order?
Suppose you can afford at most 3 ingredients • How many different types can you order? • You can order an omelet w/ 0, or 1, or 2, or 3 items and there are 10 items to choose from.
Counting problems that involve ‘_________’ or ‘_________’ sometimes are easier to solve by subtracting possibilities you don’t want from the total number of possibilities.
Subtracting instead of adding: • A theatre is having 12 plays. You want to attend at least 3. How many combinations of plays can you attend? • You want to attend 3 or 4 or 5 or … or 12. • From this section you would solve the problem using: ________________________________ • Or……
For each play you can attend you can go or not go. • So, from section 12.1 it would _________________________ • And you will not attend 0, or 1, or 2. • So:
Subtracting instead of adding: • A summer concert series has 12 different artists performing. You want to attend at least 2 of them. How many different combinations of concerts can you attend?
Subtracting instead of adding: • An amusement park has 20 different rides, and you want to go on at least 17 of them. How many different combinations of rides can you go on?
Your favorite football team is playing a 10 game season. In how many ways can they end with 5 wins and 4 losses and 1 tie?
10 people go to dinner together. In how many different ways can you order 3 chicken dinners, 2 steak dinners, and 5 lobster dinners?
There are 12 people on a committee, 8 women and 4 men. In how many ways can you elect 2 group leaders if you chose 1 man and 1 woman?
The Binomial Theorem • 0C0 • 1C0 1C1 • 2C02C12C2 • 3C03C13C2 3C3 • 4C04C14C24C34C4 • Etc…
Pascal's Triangle! • 1 • 1 1 • 1 2 1 • 1 3 3 1 • 1 4 6 4 1 • 1 5 10 10 5 1 • Etc… • This describes the coefficients in the expansion of the binomial (a+b)n
(a+b)2 = a2 + 2ab + b2 (1 2 1) • (a+b)3 = a3(b0)+3a2b1+3a1b2+b3(a0) (1 3 3 1) • (a+b)4 = a4+4a3b+6a2b2+4ab3+b4 (1 4 6 4 1) • In general…
(a+b)n = The Binomial Theorem (a+b)2 = (x-3)4 =
(a+3)5 = • 5C0a530+5C1a431+5C2a332+5C3a2b3+5C4a134+5C5a035= • 1a5 + 15a4 + 90a3 + 270a2 + 405a + 243