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NemiM [27]
3 years ago
10

Consider generating length-7 strings of lowercase letters. How many strings are there that either begin with 2 consonants or end

with 3 vowels
Mathematics
1 answer:
Semmy [17]3 years ago
7 0

Answer:

5259544316

Step-by-step explanation:

Given that:

Length of string = 7

Either begins with 2 consonants or ends with 2 vowels :

Either or :

A U B = A + B - (AnB)

Number of vowels in alphabet = 5

Number of consonants = 21

2 consonants at beginning :

First 2 consonants, then the rest could be any:

21 * 21 * 26 * 26 * 26 * 26 * 26 = 5239686816

3 vowels at the end :

First 4 letters could be any alphabet ; last 3 should be vowels.:

26 * 26 * 26 * 26 * 5 * 5 * 5 = 57122000

2 consonants at beginning and 3 vowels at the end :

21 * 21 * 26 *26 *5* 5 * 5 = 37264500

Hence,

2 consonants at beginning + 3 vowels at end 2 consonants at beginning - 2 consonants at beginning and 3 vowels At end

(5239686816 + 57122000) - 37264500

= 5259544316

Hence, number of 7 alphabet strings that begins with 2 consonants and end with 3 vowels = 5259544316

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