# Counting 11 letter words with subword "FRED" - check my answer

I'm planning for a test and also simply intend to see to it I am obtaining points appropriately.

The amount of 11 - letter 'words' created from the 26 letters of the english alphabet have the subword "FRED" Where 'words' are simply any kind of string of 11 letter words (does not need to be a word in the thesaurus)

a subword is a word inside an additional word. So an 11 letter word have to have words "FRED" in it, so FREDAABBCDEF is permitted, or XYZFREDABBC

My solution: FRED has 4 letters, so:

Choose the 4 placements to place FRED in: $c(11,4)$¢ Choose the continuing to be 7 letters (can be any kind of repeatable letters): $p(26,7)$

So we have $c(11,4) * p(26,7)$

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2022-06-07 14:34:35
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The solution that you recommend is wrong, however in greater than one means. Below is a concrete evaluation of the trouble, extensive yet I wish rather uncomplicated. Your blunder was to attempt to make use of a straightforward formula, prior to executing an evaluation of what is actually taking place.

The amount of $11$ letter words have FRED as a subword, with the letters FRED, because order, inhabiting the first $4$ placements in words? The continuing to be letters can be completed $26^7$ means, so the solution is $26^7$.

The amount of words have FRED as a subword, with the letters FRED, because order, inhabiting placements $2$ to $5$ in words? The continuing to be letters can be completed $26^7$ means, so the solution is $26^7$.

The very same holds true for words that have actually FRED inhabiting placements $3$ to $6$, $4$ to $7$, $5$ to $8$, and more approximately $8$ to $11$.

Currently it is alluring to build up, obtaining a total amount of $(8)(26^7)$. Nonetheless, this will certainly double matter, as an example, words that start with FRED, after that some pointless letter, after that FRED once more, with $2$ letters included at the end. There are numerous various other sorts of words that have 2 FRED remains in them that will certainly be double - counted.

The good news is, we do not need to bother with words that have $3$ or even more events of FRED. (If we were managing $47$ - letter words, points can get really undesirable.)

Allow us figure out the amount of double - counted words there are. We will certainly do it gradually, and afterwards in a slicker means.

The amount of "two FRED" words have the first one at the start? The 2nd one can after that start in any one of placements $5$ to $8$, so $4$ placements in all. For each and every of thse selections, the continuing to be "useless" letters can be positioned in $26^3$ means, for a total amount of $(4)(26^3)$.

The amount of "two FRED" words have the first FRED beginning at placement $2$? The very same thinking as in the paragraph over programs that there are $(3)(26^3)$ such words.

The amount of have the first FRED beginning at placement $3$? Plainly $(2)(26^3)$. The amount of with the first FRED beginning at placement $4$? Plainly $(1)(26^3)$. Which is all.

So in complete the variety of "two FRED" words is $(10)(26^3)$.

Remember that with some double - checking, we obtained a total amount of $(8)(26^7)$. Remove the double checking by deducting $(10)(26^3)$ and also we get $$(8)(26^7) -(10)(26^3)$$

A slicker means: We define an additional means of counting the "two FRED" words. Take some Scrabble floor tiles, and also adhesive them with each other to create 2 duplicates of words "FRED." Think of these as 2 unique huge letters. To make a "two FRED" word, we take our $2$ huge letters, and also $3$ average letter, and also placed them straight. So we are making an unusual $5$ - letter word. The placements of both huge letters can be picked in $\binom{5}{2}$ means. For each and every such selection, the various other letters can be completed $26^3$ means, for a total amount of $\binom{5}{2}(26^3)$. This is specifically the matter of $(10)(26^3)$ that we entered a slower means previously.

Charming, yet possibly greater than a little unsafe. The even more slow-paced strategy possibly allows us preserve better control over what is taking place.

Comment: It is all also very easy to count in a probable manner in which becomes wrong. As a partial check, one could attempt to make use of the very same thinking on say $5$ - letter words, and also MO as opposed to FRED. Make use of the sort of "general" thinking stated over, and also compare to a thorough hand - matter.

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2022-06-07 15:08:07
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You double - matter words. This is a really constant mistake in this type of troubles, where you deal with some placements to have a details value, and also pick openly the values for the various other placements. The trouble is that those various other placements might additionally have the details value, therefore you'll double count the remedy.

Below is a straightforward instance: Count all binary words which contain at the very least one event of 1. There are 3 of them, certainly: 11,01,10 (yet not 00). Nonetheless, your method returns 4: You pick an area in which to position "1" (2 selections: either first or last letter) and afterwards you pick the value of the various other placement (2 selections: 0 or 1). It is very easy to see that you count "11" two times.

Generally such troubles are addressed making use of the Inclusion-Exclusion principle.

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2022-06-07 15:00:27
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