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6.1 Use Combinations and Binomial Theorem

6.1 Use Combinations and Binomial Theorem. 378 What is a combination? How is it different from a permutation? What is the formula for n C r ?. In the last section we learned counting problems where order was important.

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6.1 Use Combinations and Binomial Theorem

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  1. 6.1 Use Combinations and Binomial Theorem 378 What is a combination? How is it different from a permutation? What is the formula for nCr?

  2. 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

  3. A Combination is a selection of “r” objects from a group of “n” objects where order is not important

  4. 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:

  5. For instance, the number of combinations of 2 objects taken from a group of 5 objects is 2

  6. 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

  7. 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:

  8. How many 7 card hands are possible? • How many of these hands have all 7 cards the same suit?

  9. Theater William Shakespeare wrote 38 plays that can be divided into three genres. Of the 38 plays, 18 are comedies, 10 are histories, and 10 are tragedies. You can choose 2 of the 18 comedies and 1 of the 10 tragedies. So, the number of possible sets of plays is: How many different sets of exactly 2 comedies and 1 tragedy can you read? 18! 10! 18C2 10C1 = 16! 2! 9! 1! 10 9! 18 17 16! = 9! 1 16! 2 1 = 153 10 = 1530

  10. You can read 0, 1, 2, or 3 plays. Because there are 38 plays that can be chosen, the number of possible sets of plays is: 38C0+ 38C1+ 38C2+38C3 = 1 + 38 + 703 + 8436 How many different sets of at most 3 plays can you read? = 9178

  11. When finding the number of ways both an event A and an event B can occur, you multiply. • When finding the number of ways that an event A ORB can occur, you +.

  12. 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?

  13. Suppose you can afford at most 3 ingredients • How many different types can you order? • You can order an omelet with 0, or 1, or 2, or 3 items and there are 10 items to choose from.

  14. Counting problems that involve ‘at least’ or ‘at most’ sometimes are easier to solve by subtracting possibilities you don’t want from the total number of possibilities.

  15. 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……

  16. For each play you can attend you can go or not go. • So, like section 12.1 it would be 2*2*2*2*2*2*2*2*2*2*2*2 =212 • And you will not attend 0, or 1, or 2. • So:

  17. Basketball During the school year, the girl’s basketball team is scheduled to play 12 home games. You want to attend at least 3 of the games. How many different combinations of games can you attend? SOLUTION Of the 12 home games, you want to attend 3 games, or 4 games, or 5 games, and so on. So, the number of combinations of games you can attend is: 12C3+12C4+12C5+………..+12C12

  18. (12C0+12C1+12C2) 212– = 4096 – (1 + 12 + 66) Instead of adding these combinations, use the following reasoning. For each of the 12 games, you can choose to attend or not attend the game, so there are 212 total combinations. If you attend at least 3 games, you do not attend only a total of 0, 1, or 2 games. So, the number of ways you can attend at least 3 games is: = 4017

  19. 6.1 Assignment Day 1 • Page 382, 3-18 all, 38-40 all

  20. 6.1 Use Combinations and Binomial Theorem Day 2 What is Pascal’s Triangle? What is the Binomial Theorem? How is Pascal’s Triangle connected to the binomial theorem?

  21. Pascal’s Triangle • If you arrange the values of nCr in a triangular pattern in which each row corresponds to a value N, you get what is called Pascal’s triangle. Pascal’s triangle is named after the French mathematician Blaise Pascal (1623-1662)

  22. The Binomial Theorem • 0C0 • 1C0 1C1 • 2C02C12C2 • 3C03C13C2 3C3 • 4C04C14C24C34C4 • Etc…

  23. 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

  24. (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…

  25. (a+b)n (n is a positive integer)= • nC0anb0 + nC1an-1b1 + nC2an-2b2 + …+ nCna0bn • =

  26. (a+3)5 = • 5C0a530+5C1a431+5C2a332+5C3a233+5C4a134+5C5a035= • 1a5 + 15a4 + 90a3 + 270a2 + 405a + 243

  27. (a+3)5 = • 1a530+5a431+10a332+10a233+5a134+1a035 • 1a5 + 15a4 + 90a3 + 270a2 + 405a + 243 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1

  28. School Clubs The 6 members of a Model UN club must choose 2 representatives to attend a state convention. Use Pascal’s triangle to find the number of combinations of 2 members that can be chosen as representatives. SOLUTION Because you need to find6C2, write the 6th row of Pascal’s triangle by adding numbers from the previous row.

  29. n = 5(5th row) 1 5 10 10 5 1 n = 6(6th row) 1 6 15 20 15 6 1 6C0 6C1 6C2 6C3 6C4 6C5 6C6 ANSWER The value of 6C2 is the third number in the 6th row of Pascal’s triangle, as shown above. Therefore, 6C2 = 15. There are 15 combinations of representatives for the convention.

  30. (x2+ y)3 = 3C0(x2)3y0+ 3C1(x2)2y1+ 3C2(x2)1y2+ 3C3(x2)0y3 Use the binomial theorem to write the binomial expansion. = (1)(x6)(1) + (3)(x4)(y) + (3)(x2)(y2) + (1)(1)(y3) = x6 + 3x4y + 3x2y2 + y3

  31. =10C0(3x)10(2)0 + 10C1(3x)9(2)1 + . . . + 10C10(3x)0(2)10 (3x + 2)10 10C6(3x)4(2)6 = (210)(81x4)(64) ANSWER The coefficient of x4 is 1,088,640. Find the coefficient of x4 in the expansion of (3x + 2)10. SOLUTION From the binomial theorem, you know the following: Each term in the expansion has the form 10Cr(3x)10 – r(2) r. The term containing x4 occurs when r = 6: = 1,088,640x4

  32. (x –3)7 7C0(x)7(–3)0 + 7C1(x)6(–3)1 + . . . + 7C7(x)0(–3)7 = = 189x5 7Cr(x)7 – r(–3)r = (21)(x5)(9) ANSWER The coefficient of x5 is 189. Find the coefficient of x5 in the expansion of (x – 3)7. SOLUTION From the binomial theorem, you know the following: Each term in the expansion has the form 7Cr(x)7 – r(–3)r. The term containing x5 occurs when r = 2:

  33. (2x + 5)8 8C0(2x)8(5)0 + 8C1(2x)7(5)1 + . . . + 8C8(2x)0(5)8 = = 1,400,000x3 8C5(2x)3(5)5 = (56)(8x3)(3125) ANSWER The coefficient of x3 is 1,400,000. Find the coefficient of x3 in the expansion of (2x + 5)8. SOLUTION From the binomial theorem, you know the following: Each term in the expansion has the form 8Cr(2x)8 – r(5)r. The term containing x3 occurs when r = 5:

  34. What is Pascal’s Triangle? An arrangement of the values of nCrin a triangular pattern. • What is the Binomial Theorem? A formula that gives the coefficients in the raising of (a+b) to a power. • How is Pascal’s Triangle connected to the binomial theorem? Pascal’s Triangle gives the same coefficients as the Binomial Theorem.

  35. Assignment 12.2 • Page 382 20-30even, 32-34 all,48-50 all

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