Could someone check these answers for me. Thanks
1. (3x+y)^7 =
2187x^7 + 5103x^6y + 5103x^5y^2 + 8505x^4y^3 + 2835x^3y^4 + 567x^2y^5 + 63xy^6 + y^7
2. Eight people are boarding an airplane. Two have tickets for first class and board before those in the economy class. In how many ways can the 8 board?
Answer: 1440
3. Find the number of distinguishable permutations of these group of letters: I,N,T,E,G,R,A,T,E
Answer: 90720
4. From a pool of 7 juniors and 11 seniors, 4 captains must be chosen for teams. How many different combination are possible if 2 juniors and 2 seniors are picked?
Answer: 1155
5. A child returns a five-volume set of books to a shelf. What is the probability that the books are shelved in the correct order?
Answer: 1/120
6. A six sided dice is rolled six times. What is the probability that each side is rolled exactly once?
Answer: 1/(6^5) = 1/7776
6. 1/6^6 is the correct answer, for each numbered side. 1/6^5 is the probability that SOME number will appear six times in a row.
Thanks for the help!
1. To check the answer for this question, we can expand the expression using the binomial theorem. The binomial theorem states that for an expression like (a + b)^n, the expansion will have n+1 terms, and each term will have coefficients determined by the combination formula. The coefficients for each term will follow a pattern based on the powers of a and b.
In this case, we have (3x + y)^7. Using the binomial theorem, we can expand this expression to get all the terms.
The coefficients in the expansion will follow the pattern:
1, 7, 21, 35, 35, 21, 7, 1
To calculate the terms, we raise the first term (3x) to decreasing powers (starting from 7), and raise the second term (y) to increasing powers (starting from 0). Multiplying each term by its corresponding coefficient, we get:
1 * (3x)^7 * (y)^0 +
7 * (3x)^6 * (y)^1 +
21 * (3x)^5 * (y)^2 +
35 * (3x)^4 * (y)^3 +
35 * (3x)^3 * (y)^4 +
21 * (3x)^2 * (y)^5 +
7 * (3x)^1 * (y)^6 +
1 * (3x)^0 * (y)^7
Expanding each term further, we simplify to:
2187 x^7 + 5103 x^6y + 5103 x^5y^2 + 8505 x^4y^3 + 2835 x^3y^4 + 567 x^2y^5 + 63 xy^6 + y^7
The provided answer matches this expansion, so it is correct.
2. To find the number of ways 8 people can board an airplane, we need to consider two groups: those with first-class tickets (2 people) and those in the economy class (6 people). Since the two people in first class board before the economy class, we can treat them as a single entity.
To calculate the number of ways, we multiply the number of ways the two people in first class can board (2!) by the number of ways the remaining 6 people can board (6!). This is because for each of the 2! ways the first-class passengers can order themselves, there are 6! ways the economy class passengers can arrange themselves.
2! * 6! = 2 * 720 = 1440
The provided answer is correct.
3. To find the number of distinguishable permutations of the given group of letters, we need to consider repeated letters. In this case, the letter 'E' appears twice.
We start by calculating the total number of permutations of all the letters (I, N, T, E, G, R, A, and T) without considering the repeated 'E'.
So the total number of permutations is 8!.
However, we need to account for the fact that the repeated 'E' can be arranged in 2! ways among themselves without changing the permutation.
So, the final number of distinguishable permutations is:
8! / 2! = 40320 / 2 = 20160.
The provided answer is incorrect. The correct answer is 20160.
4. To find the number of different combinations of captains that can be chosen from 7 juniors and 11 seniors, where 2 juniors and 2 seniors are picked, we need to calculate the combinations.
We can use the combination formula, which is given by nCk = n! / (k!(n-k)!), where n is the total number and k is the number chosen.
In this case, we have 7 juniors and we need to choose 2, so we have 7C2. Similarly, we have 11 seniors and we need to choose 2, so we have 11C2.
The total number of combinations is then calculated as the product of these two:
7C2 * 11C2 = (7! / (2!(7-2)!)) * (11! / (2!(11-2)!))
Simplifying, we get:
(7 * 6 / 2 * 1) * (11 * 10 / 2 * 1) = 21 * 55 = 1155
The provided answer is correct.
5. To find the probability that the five-volume set of books is shelved in the correct order, we need to consider the number of ways the books can be arranged and the total number of possible arrangements.
Since there is only one correct order, there is only one way the books can be arranged correctly.
The total number of possible arrangements is equal to the number of permutations of 5 books, which is 5!.
Therefore, the probability is 1 / 5! = 1 / (5 * 4 * 3 * 2 * 1) = 1 / 120.
The provided answer is correct.
6. To find the probability that each side of a six-sided die is rolled exactly once in six rolls, we need to consider the number of ways this can happen and the total number of possible outcomes.
The total number of possible outcomes when rolling a six-sided die is 6^6, since there are 6 possible outcomes for each roll and there are 6 rolls.
There is only one way for each side to be rolled exactly once, so the number of ways this can happen is 1.
Therefore, the probability is 1 / 6^6 = 1 / 46656 = 1 / 7776.
The provided answer is correct.