Top 10 How Many Different Linear Arrangements Are There Of The Letters 126 Most Correct Answers

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How many different arrangements can be formed from the letters in tooth?

We can arrange the letters in the word TOOTH in 30 different orders.

How many arrangements are possible for the 3 letters?

26⋅26⋅26=263=17576. If you want the letters to be unique, the calculation changes slightly.

How do you calculate arrangements of letters?

The second space can be filled by any of the remaining 3 letters. The third space can be filled by any of the 2 remaining letters and the final space must be filled by the one remaining letter. The total number of possible arrangements is therefore 4 × 3 × 2 × 1 = 4!

How many arrangements are there in the letters of the word math?

There are 24 different ways to arrage the letters in the word math .

How many different ways can we arrange the letters Abeacdeaaccb?

Total possible arrangement of letters a b c d is 24.

How many different letter arrangements can be formed from the letters pepper?

(2!) =60 possible letter arrangements of the letters PEPPER.

How many 3 letter combinations are possible from ABCD?

Notice that a three-letter word has all different letters, two letters that are the same and one that is different, or all three letters the same. Without any restrictions on the number of repetitions, we found 216 three-letter words.

How many combinations can you make with 4 letters?

Any choice of four distinct letters gives rise to 24 words. Thus the list of words is 24 times too large. Hence the number of combinations of 4 letters from 8 letters is 1680/24 = 70.

How many combinations are there with 26 letters?

Answer and Explanation: The number of possible combinations that are possible with 26 letters, with no repetition, is 67,108,863.

What is letter arrangement?

n. A sorting technique that orders entries based on each character in a heading without regard for spaces or punctuation.

How many different arrangements of the letters in the word tomorrow can be made?

I know TOMORROW can be arranged in 8! 3! 2! =3360 ways.

How do you know how many arrangements a word has?

Therefore, we will use Permutations to ‘arrange’ the 6 letters of the given word. Thus, the formula is nPr=n! (n−r)! Where, n is the total number of letters and r represents the number of letters to be arranged, i.e. 6 in each case.

How many arrangements can be made with the letters of the word MATHEMATICS ‘? In how many of them vowels are together?

1 Answer. James L. Mathematic can be arranged in 453,600 different ways if it is ten letters and only use each letter once. Assuming all vowels will be together 15,120 arrangements.

How many different ways can the letters of the word MATHEMATICS be arranged such that the vowels must always come together?

Thus, we have MTHMTCS (AEAI). Number of ways of arranging these letters =8! / ((2!)( 2!)) = 10080.

How many three letter arrangements can be formed from the letters of the word challenges?

How many different arrangements are possible if 3 letters are randomly selected from the word CHALLENGE and arranged into ‘words’? but the answer is 246. Your calculation gives the number of distinguishable arrangements of all of the letters in the word CHALLENGE.

What is 9c9 in math?

Plugging in our numbers of n = 9 and r = 9, we get

9C9 = 9!

How many ways can you arrange 2 letters from the word square?

The number of 2-letter words is (62)⋅2! =30.

How many different 5 letter arrangements are there of the letters in the word moose?

2. How many different 5-letter arrangements are there of the letters in the word moose? Explanation Permutation: 5P5/2! The o repeats twice. =(5•4•3•2•1)/(2•1) = 120/2 = 60
Choose: 120 60 30 20

Video 3 : Linear arrangement of letters
Video 3 : Linear arrangement of letters


probability – Arrangement probabilities – Mathematics Stack Exchange

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how many different linear arrangements are there of the letters

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How many ways can the word math be arranged? | Socratic

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  • Most searched keywords: Whether you are looking for How many ways can the word math be arranged? | Socratic Updating There are 24 different ways to arrage the letters in the word “math”. Think about it like this: If you pick any letter (“m, a, t,” or “h”) for the first “letter slot” in the word, there are four different choices. Then, for the next “slot”, you have three other letters to choose from to put in there, so that triples the combinations. That’s already 4*3 possible ways, or 12. For the third slot, you only have two other letters to choose from, so that doubles the combinations, making it 24. Finally, for the last slot, you only have one letter to choose from, so that multiplies the combinations by 1, keeping it at 24. (Also, if you understand Python, you can easily make a computer figure this out for you using two lines: But for school purposes, that’s probably considered cheating.)
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Q. 1.1 How many different linear arrang… [FREE SOLUTION] | StudySmarter

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Short Answer

Part (a) Step 1 Find the no of linear arrangements of the letters A B C D E F such that A and B are next to each other

Part (b) Step 1 Find the no of linear arrangements of the letters A B C D E F such that A is before B

Part (c) Step 1 Find the no of linear arrangements of the letters A B C D E F such that A is before B and B is before C

Part (d) Step 1 Find the no of linear arrangements of the letters A B C D E F such that A is before B and C is before D

Part (e) Step 1 Find the no of linear arrangements of the letters A B C D E F such that A and B are next to each other and C and D are also next to each other

Part (f) Step 1 Find the no of linear arrangements of the letters A B C D E F such that E is not last in line

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SOLVED:How many different linear arrangements are there of the letters A, B, C; D_ E, F for Which (a) A and B are next t0 each other; (b) A is before B; A is before B and B is before C; 8 A is before B and C is before D; A and B are next t0 each other and C and D are also next t0 each other; (f) E is not last in line?

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  • Most searched keywords: Whether you are looking for SOLVED:How many different linear arrangements are there of the letters A, B, C; D_ E, F for Which (a) A and B are next t0 each other; (b) A is before B; A is before B and B is before C; 8 A is before B and C is before D; A and B are next t0 each other and C and D are also next t0 each other; (f) E is not last in line? The total number of arrangements are six factorial. It can be either of B or after be. Therefore, this arrangement is equal to six factorial dived by two … VIDEO ANSWER:in the question. It is given that there are six letters such as a, B, C, D E and F. And we have to determine how many different linear arrangements are possible. In the first case it is given that A and B are next to each other. If n we are next to each other or together, then it will be one unit. Therefore this will be A B, then C, D E and F. This arrangement can be done by five factorial ways. Here it is four plus one. A B will be one unit. Then maybe themselves can be arranged by two factorial ways. Therefore, total arrangement is equal to five factorial into two factorial which is equals to five into four, into three into two. Into one into one into two, which is equals to 1 20 into two, which is equals to 2 40. Therefore, the answer for a parties to 40 in the B part. The condition is given that is before B. The total number of arrangements are six factorial. It can be either of B or after be. Therefore, this arrangement is equal to six factorial divided by two factorial which is equal to 6 70 divided by two, which is equal to 3 60 in the sea part. It is given that A is before B and blogs before. See we have already discussed that total number of arrangements are six. Therefore abc can be arranged by three factorial ways. Therefore, total arrangement is equal do six factorial divided by three factorial. That is arrangement of this ACC which is equals to into three factorial divided by three factorial. Here’s three factorial get canceled and is equal to six and 25 and four which is equal to 1 20 in the deep part. It is given that A’s before B and c’s before deep in the A part we have seen that the total number of arrangements when A is before B. Is it quiet dude? 3 60. And in the sea part we have seen that arrangement when C is before the is equal do mm hmm. 3 60 divided by two which is equals to 18. Therefore, the answer for the party’s 1 80 in the e bike. The problem is and we are next to each other and cnd are also next to each other. Therefore, this arrangement can might be a B and cd. There are four factorial which is equal to four units. And this can be done by two factorial into two factorial. Therefore, total arrangement is equal to four factorial into two factorial into two factorial which is equals to 4321 into two into two which is equal to 96. In the last case, that is in the f part. The condition is given that is not last in the line. The total number of arrangements when is fixed at the last place is equal to suppose there are we have seen that A B, c d E and f r. The 56 numbers. Therefore, this arrangement might be 12345, and here it is E, which is equals to five factorial, and total number of arrangements are six factorial. Therefore, the number of arrangements in which is in the last place. Is it clear to six factorial minus 55 factorial, Which is equals to 7 20 minus 1 20 which is equal to 600. Therefore, the answer for f parties 600.
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SOLVED:How many different linear arrangements are there of the letters A, B, C; D_ E, F for Which (a) A and B are next t0 each other; (b) A is before B; A is before B and B is before C; 8 A is before B and C is before D; A and B are next t0 each other and C and D are also next t0 each other; (f) E is not last in line?
SOLVED:How many different linear arrangements are there of the letters A, B, C; D_ E, F for Which (a) A and B are next t0 each other; (b) A is before B; A is before B and B is before C; 8 A is before B and C is before D; A and B are next t0 each other and C and D are also next t0 each other; (f) E is not last in line?

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$\begingroup$

I worked out each part, but I am not really sure about my answers after part a.

How many different linear arrangements are there of the letters A B C D E F for which:

a. A and B are next to each other? 2!5! = 240

I grouped A and B as “one element” so now there’s 5 elements, so 5! ways to arrange them, multiplied by the 2! ways to arrange AB (either AB or BA)

b. A is before B?

correct answer: 6!/2

1st attempt:

5!+4!+3!+2!+1!= 153

The way I pictured this problem is:

A _ _ _ _ _ 5! ways for B to be arranged

X A _ _ _ _ 4! ways for B to be arranged

X X A _ _ _ 3! ways for B to be arranged

X X X A _ _ 2! ways for B to be arranged

X X X X A _ 1! ways for B to be arranged

c. A is before B and B is before C?

1st attempt:

$5!4!(3!)^22!1!$

5! ways to choose letter A

4! ways to choose letter B

3! ways to choose letter C

3! ways to choose letter 4

2! ways to choose letter 5

1! way to choose letter 6

d. A is before B and C is before D?

1st attempt:

(I am not at all confident about this answer)

5!4! ways for A to be before B, then 3!2! ways for C to be before D, multiplied by 2! because we could do the reverse (5!4! ways for C to be before D, then 3!2! ways for A to be before B

e. A and B are next to each other and C and D are also next to each other? = $4!2!2!$

So I am grouping AB as “one element” and CD as one element”, so there are now 4 element.

4! ways to arrange them

2! ways to arrange AB

2! ways to arrange CD

f. E is not the last in line? = $(5!)^24!3!2!1!$

5! ways to choose E

5! ways to choose letter 2

4! ways to choose letter 3

3! ways to choose letter 4

2! ways to choose letter 5

1! ways to choose letter 6

How many possible 3 character combinations using letters?

How many possible 3 character combinations using letters?

Q1: How many combinations can be using 3 alphabet letters (from A to Z)? Note that string doesn’t have to be unique, it can be eg. RPP or AZZ

sol: You can make three independent choices, one for each of the three letters. For each choice, you have 26 options (the letters in the alphabet). So the total number of combinations is

26 ⋅ 26 ⋅ 26 = 26 3 = 17576.

If you want the letters to be unique, the calculation changes slightly. You still have 26 options for the first choice, but for the second choice there are now only 25 options available (all letters except the one you already chose), and for the third choice there are 24 options available (all letters except the two you already chose). So this gives you:

26 ⋅ 25 ⋅ 24 = 15600.

TASK FOR YOU

Q1: How many possible 7 characters combination using numbers and letters(alphabets)?

Permutations and Combinations

This section covers permutations and combinations.

Arranging Objects

The number of ways of arranging n unlike objects in a line is n! (pronounced ‘n factorial’). n! = n × (n – 1) × (n – 2) ×…× 3 × 2 × 1

Example

How many different ways can the letters P, Q, R, S be arranged?

The answer is 4! = 24.

This is because there are four spaces to be filled: _, _, _, _

The first space can be filled by any one of the four letters. The second space can be filled by any of the remaining 3 letters. The third space can be filled by any of the 2 remaining letters and the final space must be filled by the one remaining letter. The total number of possible arrangements is therefore 4 × 3 × 2 × 1 = 4!

The number of ways of arranging n objects, of which p of one type are alike, q of a second type are alike, r of a third type are alike, etc is:

n! .

p! q! r! …

Example

In how many ways can the letters in the word: STATISTICS be arranged?

There are 3 S’s, 2 I’s and 3 T’s in this word, therefore, the number of ways of arranging the letters are:

10! =50 400

3! 2! 3!

Rings and Roundabouts

The number of ways of arranging n unlike objects in a ring when clockwise and anticlockwise arrangements are different is (n – 1)!

When clockwise and anti-clockwise arrangements are the same, the number of ways is ½ (n – 1)!

Example

Ten people go to a party. How many different ways can they be seated?

Anti-clockwise and clockwise arrangements are the same. Therefore, the total number of ways is ½ (10-1)! = 181 440

Combinations

The number of ways of selecting r objects from n unlike objects is:

Example

There are 10 balls in a bag numbered from 1 to 10. Three balls are selected at random. How many different ways are there of selecting the three balls?

10C 3 = 10! = 10 × 9 × 8 = 120

3! (10 – 3)!3 × 2 × 1

Permutations

A permutation is an ordered arrangement.

The number of ordered arrangements of r objects taken from n unlike objects is:

nP r = n! .

(n – r)!

Example

In the Match of the Day’s goal of the month competition, you had to pick the top 3 goals out of 10. Since the order is important, it is the permutation formula which we use.

10P 3 = 10!

7!

= 720

There are therefore 720 different ways of picking the top three goals.

Probability

The above facts can be used to help solve problems in probability.

Example

In the National Lottery, 6 numbers are chosen from 49. You win if the 6 balls you pick match the six balls selected by the machine. What is the probability of winning the National Lottery?

The number of ways of choosing 6 numbers from 49 is 49C 6 = 13 983 816 .

Therefore the probability of winning the lottery is 1/13983816 = 0.000 000 071 5 (3sf), which is about a 1 in 14 million chance.

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