HW Wk 9: Combinatorics

Find the number of strings of six lowercase letters from the English alphabet, which satisfies the given conditions. Six lowercase letters contain the letters a and b

11737502 [26^6 − (25^6 + 25^6 − 24^6)]

How many permutations of the letters ABCDEFGH satisfies the given conditions? Permutation contains the strings BA and FGH.

120

Find the number of strings of six lowercase letters from the English alphabet, which satisfies the given conditions. Six lowercase letters contain the letters a and b in consecutive positions with a preceding b, with all the letters distinct.

1275120 5P(24, 4) = 5(24! / (24 - 4)!)

Find the coefficient of x^8 in (2 - x)^19.

154791936

A club has 27 members. How many ways are there to choose four members of the club to serve on an executive committee?

17550 C(27,4)=27! / 4!(27−4)! =27 · 26 · 25 · 24 / 4 · 3 · 2 · 1

Consider the statement that if n is a positive integer, then (2n2)=2(n2)+n2(22n)=2(2n)+n2 . Which of the following is the proof?

2(n2)+n2=n(n−1)+n2=2n2−n=n(2n−1)=2n(2n−1)2=(2n2)2(2n)+n2=n(n−1)+n2=2n2−n=n(2n−1)=2n(2n−1)2=(22n)

A coin is flipped where, each flip comes up as either heads or tails. How many possible outcomes are there in total if the coin is flipped 11 times?

2048 there are 2^11 total outcomes.

How many permutations of the letters ABCDEFGH satisfies the given conditions? Permutation contains the strings CAB and BED?

24

Find the coefficient of x^6y^8 in (x + y)^14.

3003

A coin is flipped where, each flip comes up as either heads or tails. How many possible outcomes contain the same number of heads and tails if the coin is flipped 14 times?

3432 C(14,7) = 14! / 7! ⋅ (14 - 7)! =(14)(13)(12)(11)(10)(9)(8) / (7)(6)(5)(4)(3)(2)(1)

Find the number of bit strings that satisfies the given conditions. The bit strings of length 7 having exactly four 1s

35 C(7, 4) = 7! / 4!(7 - 4)! = (7 ⋅ 6 ⋅ 5 ⋅ 4) / (4 ⋅ 3 ⋅ 2 ⋅ 1)

A coin is flipped where, each flip comes up as either heads or tails. How many possible outcomes contain exactly two heads if the coin is flipped 9 times?

36 C(9,2) = 9! / 2!(9 - 2)! = (9 ⋅ 8) / 2

Find the number of strings of six lowercase letters from the English alphabet, which satisfies the given conditions. Six lowercase letters contain the letters a and b, where a is somewhere to the left of b in the string, with all the letters distinct.

3825360 C(6, 2)P(24, 4) = (6! / 2!(6 - 2)!)(24! / (24 - 4)!)

A club has 27 members. How many ways are there to choose a president, vice president, secretary, and treasurer of the club, where no person can hold more than one office?

421200

The number of subsets with more than two elements that can be formed from a set of 102 elements is ≈ _____ × 10^29. (Enter the value in decimals. Round the answer to two decimal places.)

50.71

How many permutations of the letters ABCDEFGH satisfies the given conditions? Permutation contains the string ED.

5040

Find the number of bit strings that satisfies the given conditions. The bit strings of length 11 having at most four 1s

562 C(11, 4) + C(11, 3) + C(11, 2) + C(11, 1) + C(11, 0) = 330 + 165 + 55 + 11 + 1

Find the number of bit strings that satisfies the given conditions. The bit strings of length 7 having at least four 1s

64 2^7 - C(7, 3) - C(7, 2) - C(7, 1) - C(7, 0) = 128 - 35 - 21 - 7 - 1

Find the number of strings of six lowercase letters from the English alphabet, which satisfies the given conditions. Six lowercase letters contain the letter a.

64775151 26^6 - 25^6

Find the number of bit strings that satisfies the given conditions. The bit strings of length 8 having an equal number of 0s and 1s

70 C(8,4) = 8!4! ⋅(8 - 4)! = (8)(7)(6)(5) / (4)(3)(2)(1)

A coin is flipped where, each flip comes up as either heads or tails. How many possible outcomes contain at most three tails if the coin is flipped 8 times?

93 C(8, 3) + C(8, 2) + C(8, 1) + C(8, 0) = 56 + 28 + 8 + 1

A group contains n men and n women. Identify the steps used to find the number of ways to arrange n men and n women in a row if the men and women alternate? (Check all that apply.)

The correct steps to find the number of ways to arrange n men and n women in a row are as follows: 1. There are n men and n women, and all of the P(n, n) = n! arrangements are allowed for both men and women. 2. Therefore, there are exactly two possibilities: either the row starts with a man and ends with a woman, or it starts with a woman and ends with a man. 3. Hence, by the product rule, there are n!· n!· 2 = 2(n!)2 ways.

Consider the statement that if n is a positive integer, then (2n2)=2(n2)+n2(22n)=2(2n)+n2 . Select the correct step(s) to prove the given identity. (Note: We are proving this statement combinatorically, by interpreting the equality as giving, in two different ways, the number of ways of choosing two people from a set of n men and n women.) (Check all that apply.)

To choose 2 people from a set of n men and n women, either choose 2 men or 2 women or one of each sex. There are C(n, 2) ways to choose two men or two women and n ⋅ n ways to choose one of each sex. The left-hand side counts ways to do so, since simply choosing 2 people from 2n people. The right-hand side counts the number of ways to do this (by the sum rule).