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Nastasia [14]
3 years ago
6

How many five letter strings are there that start with a consonant and end with Y where repetition is not permitted

Mathematics
1 answer:
Arada [10]3 years ago
6 0

Answer: 242,880

Step-by-step explanation:

We know that:

We have 5 letters.

There is no repetition allowed.

The strings start with a consonant.

The strings end with the letter Y.

Ok, in our alphabet we have:

26 letters.

21 consonants

5 vowels.

Now, let's count the number of options for each one of the letters in the string.

First letter, this must be a consonant, initially, we can have 21 options, but this string must finish with an Y, and Y is a consonant, so it can not be in the first letter, then for the first letter we have 20 possible options.

Now we have a total of 26  - 1 - 1 = 24 letters left, where we subtracted the consonant for the first letter, and the Y for the last letter.

The options for the second letter will be 24.

The options for the third letter will be 23

The options for the fourth letter will be 22

For the last one we have a fixed value, so we have only one option.

The total number of combinations will be equal to the product between the numbers of options for each letter, this is:

C = 20*24*23*22 = 242,880 possible strings.

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