How many strings of five English letters that start with two consecutive vowels (A, E,I, O, or U) have no repeated letters

242880

Step-by-step explanation:

In the English alphabet, we have 5 vowels. To form a string of 5 English letters in which it starts with two consecutive vowels; then we have the following.

In the first space, 5 vowels are possible.

in the second space, 4 vowels are possible since there is no room for repetition of letters.

Now, the total number of ways of assigning two distinct vowels in the first two spaces = 5 × 4 = 20

From the question, since it is said that the letter starts with two consecutive vowels, it implies that starting from the third letter, it can be filled with all the remaining alphabet, be it a vowel or consonant.

Thus, the remaining alphabet = 26 - 2 = 24 alphabets to fill the 3 remaining spaces, which can be achieved in 24P3 ways.

P(24,3) = 24!/(24-3)!

P(24,3) = 24!/21!

P(24,3) = 24 × 23 × 22 × 21! / 21!

P(24,3) = 24 × 23 × 22

P(24,3) = 12144 ways

Thus, the total number of strings of five  English letters that start with two consecutive vowels without having repeated letters is:

= 5  × 4  ×  24 × 23 × 22

= 242880

 5        4        24          23       22  

there would be 365 different 4 digit numbers

step-by-step explanation:

i belive that it is 365 because i asked google

hope this !

step-by-step explanation:

d = distance between moon and sun, which in this problem is the hypotenuse of a right triangle.  

the distance y is the side opposite to angle x in the right triangle.   use the definition of the sine function from basic trigonometry.

sin(x) = y/d ⇒

d = y/sin(x)