How many strings of eight English letters are there a) that contain no vowels, if letters can be

How many strings of eight English letters are there a) that...

Key Concepts

Multiplication Principle

This principle is the backbone of counting problems in combinatorics, stating that if there are n ways to perform one task and m ways to perform another, then there are n × m total ways to perform both. In the context of string formation, it allows us to multiply the number of choices for each letter position to find the total number of possible strings under the given restrictions.

Counting with Repetition

When repetition of elements is allowed, each position in a string can be chosen independently from the entire set of valid elements for that position. This means that the same letter can be used multiple times in different positions, simplifying the counting process as each position has a fixed number of available choices regardless of the selections made in other positions.

Permutations Without Repetition

In situations where letters (or items) cannot be repeated, each selection reduces the pool of available choices for subsequent positions. The count is determined by the product of a sequence of decreasing numbers, reflecting the fact that once a letter is used, it is no longer available for selection in the following positions.

Complement Principle

This technique is used to count the number of outcomes that satisfy 'at least one' type constraints. Instead of counting the desired outcomes directly, it is often easier to calculate the total number of outcomes without the restriction and then subtract the number of outcomes that do not meet the requirement (the complement).

Position-specific Restrictions

Many counting problems involve additional constraints that affect only certain positions in a string, such as requiring that a string starts with a vowel or a specific letter. These restrictions reduce the number of choices for those positions independently from the rest of the positions, and the overall count is computed by incorporating these adjusted counts into the multiplication principle.

Transcript

0:00 Hello there.

00:01 In this exercise we need to count how many strings of english letters of length 8 exist if we put some conditions.

00:11 So the first condition that we need to consider is that there is no bowels on the string.

00:18 So no vowels but also we can repeat is allowed to repeat the letters.

00:27 So the repetition is allowed.

00:29 So we can repeat the letters.

00:34 And so let's start to count.

00:37 Okay, so we know that the alphabet, the english alphabet, contains exactly 26 charters between bowels and consonants, where five of them are bowels, and then 21 of them are consonants.

01:00 So if or a string that is, has like length 8 1 2 3 4 5 6 7 8 okay here is our string of length 8 can it is not allowed to have bowels then we we are left with only 21 consonants or charters so in each position of this string we we have 21 possible choices for characters or letters to be more precise to put on the string and even more we are we the repetition is allowed so that means that we can repeat the letters and we can in each position of the string we have 21 possibilities to put a constant there so each position has 21 possible possible letters to put on it so at the end we have 21 that repeat eight times because of the eight positions on the string and that means that we have 21 to the eighth power possible ways to to arrange these strings with no vowels now what happened if instead of allowing the repetition again we have we are not allowing the bowels but also no repetitions so if we are not allowing the repetitions, that means that in our string, again, we have eight places here.

02:47 We have our string of length eight.

02:50 But now at each position, again, we are not allowing the bowels, so we are going to work with 21 letters.

02:57 So in the first position we have 21 possibilities.

03:00 But in the second one, because we are not allowing the repetitions, the one that we put here, we cannot put it on this position.

03:08 On the string so that means that we have here 20 possible outcomes for that place in the string and then for the next we have 29 sorry and 18 and then 18 and so on until the end of the string so at the end of the string we have number 14 so you can see that we need to multiply all these numbers and this equals to 20 times 21 times 20 times 19 times 18 times 17 times 16 and times 15 and times 14 and then we need to multiply all these numbers to obtain the all the ways to arrange this and the string with no repetition there is a more elegant way to put this and it's basically you have 21 characters and you have like eight spots where you're going to to order these 21 characters right so to count this how to arrange these 21 elements on eight places eight spots or eight boxes in case that you have some bowls or items that you want to stir on boxes well the the formula is given by 21 factorial divided by 21 minus the places where we are going to allocate these 21 elements in this case these 21 letters so where we're going to put these 21 letters where in our 8 length string so here we put the number 8 factorial and that is equivalent to the number that we have put here in the upper part and the result of course is a big number you can see that we need to multiply a lot of things so in case that you need the default number i'm going to write down here 8 and 2 zeros here yeah that's the number okay let's continue now we need to count how many of these strains are if we start with a vowel and again in this case we are allowing the repetitions.

05:47 So we have repeated letters.

05:50 We can have repeated letters.

05:54 So in this case, or 8 bit string, we have here or 8 bit string, the first position should start with a vowel.

06:08 And at the beginning we saw that we have five vowels on the alphabet.

06:13 So we have five possibilities in this first place.

06:17 And because we are allowing repetitions of the letters then in the next position we have 26 possible letters to put on that place because now there is no restriction with vowels so we have the 26 possible characters of the alphabet of the english alphabet to put on this string and you can see that from here to the end we have seven positions on our string the eight the seven remaining positions of the string because the first one is a bowels so in each position of this like sub string that appear here we have 26 possible letters to put at each position.

07:00 So the possible outcomes that we have all the possible ways to arrange this string is equal to 5 times 26 to the 7 power because of the 7 places in the remaining spots of the string.

07:15 And of course the result of this is 72, 343, 541, and 640.

07:26 Okay, this is the full number.

07:29 But if you put this is enough, i think.

07:34 Okay, um, oh sorry, this is not the number.

07:40 I have a mistake here is 40, 050 and 8008.

07:50 Yeah, this is the number.

07:51 Okay.

07:52 Okay.

07:52 Now, we need to start with a vowel like in the previous part we need to start with a bubble but now we are not allowing the repetitions with no repetitions okay so if we are not allowing the repetitions we have something similar so we have here three six eight or eight bit or eight a string in the first position again we have we need to start with a bowel so here we have five possible outcomes but then in the next position we have 25 possibilities because one bowel we have put in the first place of this string so in the second one we have 25 remaining letters and for next part we have 24 and then 23 and then 22 21 20 and 19 so we need to multiply all these numbers to obtain the result but as i mentioned before we have a more relevant way to write this that is five times so in this case we have 25 elements that we need to distribute in seven places that are the seven remaining places on our string because the first one is already occupied by one by one bowel so here we have divided by 25 minus seven factory of course we can we can resolve this point on the parentheses so we have 5 times 25 factorial divided by 18 factorial and this will give you the same result as multiplying all these numbers here but it's a more compact way to write down and the result of this is let me put here the number is 12 -113 -64 -0 and three zero is here so this is the full number in case that you need it.

10:05 Great.

10:07 Now we need to count how many of these strings are if they have at least one vowel.

10:22 Okay, so we need one vowel, at least one vowel in our string.

10:28 So let's consider the following.

10:31 We have here 3, 6, 8...

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