1.how many different permutations can you make with the letters in the word s e v e n t e e n ?
2.a teacher has a set of 12 probelms to use on a math exam. the teacher makes different versions of the exam by putting 10 questions on each exam. how many different exams can the teacher make?
3.how many arrangements of the letter o l i v e can you make if each arrangement must use three letters?
1.
SEVENTEEN = EEEE NN S T V
So there is one group of 4 letters, one of 2 letters, and three of 1 letter for at total of 9 letters.
The number of permutations is
P(9,9)/(4!2!1!1!1!)
=9!/(4!2!1!1!1!)
2. Think of the two questions she can leave out. She has a choice of 12 for the first one, and 11 for the second. Thus she can make 12*11=132 different exams, if the order of the questions do not matter.
If the order of the questions matters, then we choose the ten questions one after another, which gives
12*11*10*9*8*7*6*5*4*3
=12!/2!
different exams.
3. How many choices do you have for choosing the first of three letters? (5)
How many for the second? (4)
How many for the third? (3)
How many arrangements in all?
1. To find the number of different permutations that can be made with the letters in the word "seventeen," we can use the formula for permutations of a set with repetition.
First, we need to determine the number of times each letter appears in the word. The letter 'e' appears three times, and each of the other letters ('s', 'v', 'n', 't') appears only once.
Using the formula, we calculate the number of permutations as:
P = (n!) / (x1! * x2! * ... * xn!)
where n is the total number of letters in the word and x1, x2, etc. represent the number of times each letter appears.
In this case, we have:
P = (9!) / (3! * 1! * 1! * 1! * 2!)
Simplifying further:
P = (9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1) / (3 * 2 * 1 * 1 * 1 * 2)
P = 362,880 / 12
P = 30,240
Therefore, there are 30,240 different permutations that can be made with the letters in the word "seventeen."
2. For the second question, we need to find the number of ways the teacher can select 10 questions from a set of 12 problems. This is a combination problem.
Using the formula for combinations, we calculate the number of combinations as:
C = (n!) / (r! * (n-r)!)
where n is the total number of items and r is the number of items to be selected.
In this case, we have:
C = (12!) / (10! * (12-10)!)
C = (12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1) / (10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 * 2 * 1)
C = 66
Therefore, the teacher can make 66 different versions of the exam.
3. For the third question, we need to find the number of arrangements of the word "olive" using three letters. Since each arrangement must use three letters, we can use the concept of combinations.
Using the formula for combinations, the number of combinations can be calculated as:
C = (n!) / (r! * (n-r)!)
where n is the total number of items and r is the number of items to be selected.
In this case, we have:
C = (5!) / (3! * (5-3)!)
C = (5 * 4 * 3 * 2 * 1) / (3 * 2 * 1 * 2 * 1)
C = 10
Therefore, there are 10 different arrangements of the letters in the word "olive" that can be made using three letters.
1. To find the number of different permutations that can be made with the letters in the word "seventeen," we calculate the factorial of the number of letters in the word. The factorial of a number is the product of that number and all the positive integers below it. In this case, we have 9 letters in the word "seventeen." So, we calculate 9 factorial (9!) as follows:
9! = 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 362,880
Therefore, there are 362,880 different permutations that can be made with the letters in the word "seventeen."
2. To find the number of different exams the teacher can make using a set of 12 problems with 10 questions on each exam, we need to use the concept of combinations. Since the problems are different versions, the order of the questions does not matter (unlike permutations). The formula to calculate the number of combinations is given by:
nCr = n! / (r!(n-r)!)
where n is the total number of items to choose from and r is the number of items to choose.
In this case, there are 12 problems, and the teacher selects 10 problems for each exam. So, we can calculate the combination as follows:
C(12,10) = 12! / (10!(12-10)!) = 12! / (10!2!) = (12 x 11) / (2 x 1) = 66
Therefore, the teacher can make 66 different exams using the set of 12 problems, with 10 questions on each exam.
3. To find the number of arrangements of the letters "olive" if each arrangement must use three letters, we need to use the concept of permutations. The formula to calculate permutations is given by:
nPr = n! / (n-r)!
where n is the total number of items and r is the number of items to be arranged.
In this case, there are 5 letters in the word "olive," and we need to arrange 3 letters. So, we can calculate the permutation as follows:
P(5,3) = 5! / (5-3)! = 5! / 2! = (5 x 4 x 3) / (2 x 1) = 60 / 2 = 30
Therefore, there are 30 different arrangements of the letters "olive" if each arrangement must use three letters.