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The total number of passwords with 2 vowels and 3 consonants is determined as follows:

1) Choose the 2 different vowels in

5C

2 ways

2) Choose the 3 consonants in

21C

3 ways

3) Arrange the 5 letters in 5! ways

So, the number of passwords is

21C3 * 5C2 * 5! = ( 21*20*19 ) / (1*2*3) * (5*4)/(1*2) * 120 = 1330 * 10 * 120 = 1596000.

The number of passwords with 2 adjacent vowels and 3 consonants is

determined as follows:

1) Choose the 2 different vowels in

5C

2 ways

2) Choose the 3 consonants in

21C

3 ways

3) Choose a position for the two adjacent vowels in 4! ways

4) Rearrange the adjacent vowels in 2! ways

So, the number of passwords with two adjacent vowels is

21C3 * 5C2 * 4!*2! = ( 21*20*19 ) / (1*2*3) * (5*4)/(1*2) * 48 = 1330 * 10 * 48 = 638400.

Therefore, the number of passwords without two adjacent vowels is

1596000 – 638400 = 957600.

Therefore, the answer is A.

Answer: A

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