# How many permutations of the 26 letters of the English alphabet do not contain the string make and sit?

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

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## How many ways can 26 letters be arranged?

26 letters can be arranged in 67,108,863 ways without any repetition of letters.

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## How many ways can you arrange 26 letters into 3?

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

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## How many combinations are there with 25 letters?

Answer and Explanation: The number of possible combinations that can be made with 25 numbers is 33,554,432.

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## How many combinations of 24 words are there?

Then for each of those lists of words, try all 2048 possible combinations for the 24th word. So if my math is correct then there's a worst case scenario of having to brute force (33!/(23! 10!)) *2048 = 189,565,009,920 possible combinations.

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## How many 20 letter combinations are there?

13367494538843734067838845976576 to be more precise. Approximately 1.33675×1031 ways, in fact!

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## How many combinations of 2048 words are there?

As I understand for BIP32. There are 2048 words. 12 words would give you 2048^12 combinations.

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## How many ways can ABCD be arranged?

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

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## How do you calculate possible combinations?

Combinations are a way to calculate the total outcomes of an event where order of the outcomes does not matter. To calculate combinations, we will use the formula nCr = n! / r! * (n - r)!, where n represents the total number of items, and r represents the number of items being chosen at a time.

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## How do you calculate possibilities?

In terms of the combinations equation below, the number of possible options for each category is equal to the number of possible combinations for each category since we are only making 1 selection; for example C(6,1) = 6, C(5,1) = 5 and C(3,1) = 3 using the following equation: C(n,r) = n! / ( r!( n - r)! )

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## How many ways are there to pick a subset of four different letters from the 26 letter alphabet?

Explanation: This is a permutation of 26 letters taken 4 at a time. To compute this we multiply 26 * 25 * 24 * 23 = 358,800.

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## How many combinations of 3 numbers are possible?

Therefore in that set of 720 possibilities, each unique combination of three digits is represented 6 times.

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## How many 6 letter combinations are there?

So the six letters can be a combination of 6×5×4×3×2×1 letters or 720 arrangements.

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## What is the number of ways to order the 26 letters of the alphabet so that no two of the vowels?

so total no of ways in which vowel don't occur together = 26! - 22!

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## How many permutations of the 26 letters of the English alphabet do not contain the string make and sit?

The answer is 100 – 32 = 68. How many permutations of the 26 letters of the English alphabet do not contain any of the strings fish, rat, or bird? See Venn diagram. Start with the universe: 26!

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## How many permutations of the 26 letters of the English alphabet do not contain the string make and cake A 0 B 26 20 C 26 D 26 18?

It will give you 26C1 x 26 C1 x 26C1 = 26x26x26 = 17576.

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## What is 4C2 combination?

We know that the formula used to solve the combination expressions is given by: nCr = n!/[r! (n – r)! Substituting n = 4 and r = 2 in the above formula, 4C2 = 4!/ [2!

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## How many combinations of n items are there?

The number of combinations of n distinct objects, taken r at a time is: Cr = n! / r! (n - r)! Thus, 27,405 different groupings of 4 players are possible.

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## What is 15C13?

15C13 = 15! 13!(15 - 13)! 15C13 = 1,307,674,368,000.

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## How many derangements are there?

DCAB, DCBA. there are only 9 derangements (shown in blue italics above).

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## How do you solve 5P5?

5P5 is the number of ways of picking 5 objects out of a group of 5 objects, where order matters. Whenever you select ALL of the objects and order matters, the formula for nPn is n! . Since 5! =5(4)(3)(2)(1)=120 , that answers the question at hand.

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## What is bip44?

Bitcoin Improvement Proposal (BIP) 44 defines the standard derivation path for wallets which generate Pay-to-Public-Key-Hash (P2PKH) addresses. BIP 44 also defines the prefixes to be used with associated extended keys.

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## What is a 24 word seed?

Technically speaking, these 24 words are a representation of a string of random digits called a seed, from which all the keys in your wallet are derived. The seed is used to generate your master private key, which generates the rest of your private keys. Private keys are used to generate corresponding public keys.

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## How secure is a 12-word seed?

A recovery phrase is usually generated from a specific list of 2,048 words. That means that your 12-word seed phrase has an iteration of 2,048 words. Therefore, it's almost impossible that someone could successfully guess your seed phrase.

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How many arrangements of the $26$ letters of the alphabet contain neither "math" nor "the"?

Let $S$ denote the set of all permutations of the alphabet; let $M$ denote the set of all permutations of the alphabet that contain the subsequence math; let $T$ denote the set of all permutations of the alphabet that contain the subsequence the.

We wish to count all permutations in the alphabet that contain neither the subsequence math nor the subsequence the. That is, we wish to count the number of elements in the set $M^C \cap T^C$.

Since $M^C \cap T^C = S - (M \cup T)$, $$|M^C \cap T^C| = |S| - |M \cup T|$$ By the Inclusion-Exclusion Principle, $$|M \cup T| = |M| + |T| - |M \cap T|$$ Hence, $$|M^C \cap T^C| = |S| - |M| - |T| + |M \cap T|$$ You correctly found that \begin{align*} |S| & = 26!\\ |M| & = 23!\\ |T| & = 24!\\ |M \cap T| & = 22! \end{align*} Hence, $$|M^C \cap T^C| = 26! - 23! - 24! + 22!$$

Where did you go wrong?

You added the number of sequences that do not contain the subsequence math, the number of subsequences that do not contain the word the, and the number of subsequences that contain both math and the, that is, $$|M^C| + |T^C| + |M \cap T|$$ Draw a Venn diagram as follows: Draw a rectangle that includes two overlapping circles. The rectangle represents the set $S$ of all permutations of the alphabet. Label one of the circles, say the one on the left, $M$. It represents the set of sequences that contain the subsequence math. Label the other circle $T$. It represents the set of sequences that contain the subsequence the. The region where the two circles intersect is $M \cap T$. The region inside the rectangles but outside the circles is $M^C \cap T^C$, which is what we want to count since is the set of sequences in $S$ that contain neither the subsequence math nor the subsequence the.

If we shade $M^C$ (with, say, lines from the lower left to the upper right), we shade everything in $S$ except for $M$. Thus, we have already shaded $M^C \cap T^C$ and the part of set $T$ outside $M$. If we now shade $T^C$ (with, say, lines from the upper left to the lower right), we shade everything in $M^C \cap T^C$ again and the region of $M$ outside $T$. We then shade $M \cap T$ (with, say, horizontal lines). Notice that we have now shaded $M \cup T$ once and $M^C \cap T^C$ twice. Put another away, we have shaded all of $S$ once and then shaded $M^C \cap T^C$ a second time. Consequently, what you counted is $$|M^C| + |T^C| + |M \cup T| = |S| + |M^C \cup T^C|$$ Using my answer above for $|M^C \cap T^C|$, we obtain $$|S| + |M^C \cap T^C| = 26! + 26! - 23! - 24! + 22!$$ which explains why you have an extra $26!$ term in your answer.