Am I correct? Please show me how if I am wrong

1.How many 6 digit numbers can be made using each of the digits {8,8,9,5,5,6} exactly once? 6!=702

2.How many more arrangements are there from all of the letters of the word MARMALADE than there are from all of the letters of the word MARINADE? 9!-8!=322560

3.Using all of the digits {1,1,1,2,2,3,3}, how many positive seven-digit integers can be written that begin and end in 3? 3x5x5x5x5x5x3=28125

4.In how many ways can 3 yellow beads, 4 blue beads and 2 black beads be strung along a straight wire if all of the beads are identical except for color? (Don't know)

5.In how many ways can all of the letters of the word CANADA be arranged? Since there is 3 a's it is divied by 3! 6!/3!=120

6.In how many ways can all of the letters of the word MOOSOMIN be arranged? 2 m's and 3 o's 8!/2!x3!

7.Using all of the digits {1,1,1,2,2,3,3}, how many positive seven-digit integers can be written that begin and end in 1? 1x5x5x5x5x5x1=3125

8.In how many different orders can a soccer team win 4 games and lose 3 games in a best-of-seven playoff series? 7!/4!x3!=20160

1. 6!/(2!2!)

2.(9!/(2!3!)-8!/2!
3.3x(5!/3!2!)x3
4.9!/3!4!2!
5. 3! 6!/3!
6.8!/2!x3!
7.1x(5!/3!2!)X1
8.7!/4!3!

#1 There are two 8's, so while there are 6! permutations of the six digits, in any one of them, the two 8's can be swapped, and it's still the same arrangements. Same for the 5's. So, there are only 6!/(2!2!) = 180 distinguishable arrangements.

The same caveat applies to all the other problems. You need review your text section and rethink your answers.

1. 180

2. 10080
3. 94. 0
4. 2,903,040
5. 120
6. 3360
7. 10
8. 35

Let's go through each question and see if the provided answers are correct:

1. How many 6 digit numbers can be made using each of the digits {8,8,9,5,5,6} exactly once?
To solve this, we need to use the concept of permutations. Permutations calculate the number of ways to arrange a set of objects. In this case, we have 6 distinct digits, so the number of 6-digit numbers that can be formed is 6! (6 factorial), which is equal to 6 x 5 x 4 x 3 x 2 x 1 = 720.

The correct answer is 720, not 702.

2. How many more arrangements are there from all of the letters of the word MARMALADE than there are from all of the letters of the word MARINADE?
To find the difference in the number of arrangements, we need to calculate the total number of arrangements for each word. The word MARMALADE has 9 letters, so there are 9! (9 factorial) arrangements possible. Similarly, the word MARINADE has 8 letters, so there are 8! arrangements possible.

To find the difference, we subtract the number of arrangements for MARINADE from MARMALADE: 9! - 8! = 362,880 - 40,320 = 322,560.

The provided answer of 322,560 is correct.

3. Using all of the digits {1,1,1,2,2,3,3}, how many positive seven-digit integers can be written that begin and end in 3?
To solve this, we need to count the number of permutations with repetition. Since the digits 1 and 2 are repeated three times, we have 3 choices for each of them, but only 1 choice for the digit 3. For the first digit, we have 3 choices, and for the last digit, we have only 1 choice, so the total number of seven-digit integers is: 3 x 3 x 3 x 3 x 3 x 3 x 1 = 81.

The provided answer of 28,125 is incorrect. The correct answer is 81.

4. In how many ways can 3 yellow beads, 4 blue beads, and 2 black beads be strung along a straight wire if all of the beads are identical except for color?
To solve this, we can use the concept of combinations. The formula for combinations is n! / (r1! * r2! * ... * rk!), where n is the total number of objects and r1, r2, ..., rk are the repetitions of each object.

In this case, we have a total of 9 beads: 3 yellow, 4 blue, and 2 black. Using the combination formula, the number of ways to arrange them is 9! / (3! * 4! * 2!) = 1680.

From what has been provided, there is no answer given for this question.

5. In how many ways can all of the letters of the word CANADA be arranged?
To solve this, we need to calculate the number of permutations, taking into account the repeated letters. The word CANADA has 6 letters, but the letter 'A' is repeated 3 times.

To find the number of arrangements, we divide the total number of permutations (6!) by the number of permutations of the repeated letters (3!): 6! / 3! = 720 / 6 = 120.

The provided answer of 120 is correct.

6. In how many ways can all of the letters of the word MOOSOMIN be arranged?
Similar to question 5, we need to calculate the number of permutations while considering the repeated letters. The word MOOSOMIN has 8 letters, with 'M' repeated twice and 'O' repeated three times.

To find the number of arrangements, we divide the total number of permutations (8!) by the number of permutations of the repeated letters: 8! / (2! * 3!) = 40,320 / (2 * 6) = 5,040.

The provided answer of 8!/2! * 3! = 5,040 is correct.

7. Using all of the digits {1,1,1,2,2,3,3}, how many positive seven-digit integers can be written that begin and end in 1?
Similar to question 3, we need to count the number of permutations with repetition. Since the digits 2 and 3 are repeated twice, we have 3 choices for each of them, and we have only 1 choice for the digit 1 at the beginning and end.

The total number of seven-digit integers is: 1 x 3 x 3 x 3 x 3 x 3 x 1 = 243.

The provided answer of 3 x 5 x 5 x 5 x 5 x 5 x 3 = 31,125 is incorrect. The correct answer is 243.

8. In how many different orders can a soccer team win 4 games and lose 3 games in a best-of-seven playoff series?
To solve this, we can use the concept of combinations. Since a soccer team must win 4 out of the 7 games, we can calculate the number of combinations: 7! / (4! * 3!) = 35.

The provided answer of 7! / (4! * 3!) = 20,160 is correct.

So, out of the provided answers, question 1, 2, 5, 6, and 8 are correct, while questions 3, 4, and 7 have incorrect answers.