Problem Solving

[GMAT math practice question]

Each letter of a 5-letter password must be a different letter of the alphabet. Each password must contain 2 vowels and 3 consonants, and the vowels must not be adjacent to each other. How many different passwords of this form can be made?

A. 957600
B. 128000
C. 159600
D. 256000
E. 720000 $$$$ $$$$

=>

The total number of passwords with 2 vowels and 3 consonants is determined as follows:
1) Choose the 2 different vowels in 5C2 ways
2) Choose the 3 consonants in 21C3 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 5C2 ways
2) Choose the 3 consonants in 21C3 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

Max@Math Revolution wrote:[GMAT math practice question]

Each letter of a 5-letter password must be a different letter of the alphabet. Each password must contain 2 vowels and 3 consonants, and the vowels must not be adjacent to each other. How many different passwords of this form can be made?

A. 95760
B. 128000
C. 159600
D. 256000
E. 720000

From the 5 slots in the sequence, the number of ways to choose 2 slots for the 2 vowels = 5C2 = (5*4)/(2*1) = 10.
Of these 10 options, the following 4 combinations are composed of adjacent slots and thus are not allowed:
Slots 1 and 2
Slots 2 and 3
Slots 3 and 4
Slots 4 and 5
Thus:
Number of slot options for the 2 vowels = (total ways to choose 2 slots) - (number of ways to choose 2 adjacent slots) = 10-4 = 6.

In the 26-letter alphabet are 5 vowels -- a, e, i, o, u -- and 21 consonants.
Thus:
Number of options for the first vowel = 5. (Any of the 5 vowels.)
Number of options for the second vowel = 4. (Any of the 4 remaining vowels.)
Number of options for the first consonant = 21. (Any of the 21 consonants.)
Number of options for the second consonant = 20. (Any of the 20 remaining consonants.)
Number of options for the third consonant = 19. (Any of the 19 remaining consonants.)

To combine the options in blue, we multiply:
6*5*4*21*19*20 ≈ 120*20*20*20 = 120*8000 = 960000.

The OA is incorrect.
It should be 957,600.

The total number of passwords with 2 vowels and 3 consonants is determined as follows:
1) Choose the 2 different vowels in 5C2 ways
2) Choose the 3 consonants in 21C3 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 = 133 * 10 * 120 = 159600.

The number of passwords with 2 adjacent vowels and 3 consonants is
determined as follows:
1) Choose the 2 different vowels in 5C2 ways
2) Choose the 3 consonants in 21C3 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 = 133 * 10 * 48 = 63840.

Therefore, the number of passwords without two adjacent vowels is
159600 - 63840 = 95760.

Therefore, the answer is A.

Answer: A

The values in red are incorrect.
(21*20*19)/(1*2*3) = 1330.