A 5-letter password must consist of different letters of the alphabet

Question

A 5-letter password must consist of different letters of the alphabet, with 2 vowels and 3 consonants. If the vowels must not be adjacent to each other, how many such passwords can be created?

A. 95,760
B. 159.600
C. 256.000
D. 720,000
E. 957,600

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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?

VivekSri · Answer

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Aryaa03 · Answer

Sol:
 IMO-Option-E

Total no. of vowel : 5 , out of which 2 are to be selected : 5C2 = 10 
Total no. of consonants: 21 , out of which 3 are to be selected : 21C3 = 1330

Let, x, No. of ways in which no two vowel are adjacent : Total no. of ways in which 5 selected can be arranged - Total arrangement considering with both vowel adjacent

a) Total no. of arrangement possible = 5! =120
b) Consider V1-V2 as one entity and thus, total arrangement is 4!*2 = 48 (multiplied by two as there are two ways in which V1 and V2 can be arranged)

Therefore, total no. of arrangements possible = 1330*10*(5! - 4!*2) = 13300*(120-48) = 957600

Archit3110 · Answer

total ways to place 2 vowels and 3 consonants
5*4*21*20*19 ; 159600
VVCCC ; 5!/3!* 2! ; 10 arrangements possibles are 10

let Vowels be placed together 
XCCC ; 4!/3! ; 
10-4 ; 6 ways vowels are not together
159600*6 ; 957600
option E

samarpan.g28 · Answer

1. There are 5 vowels. Select any 2 in 5C2 ways.
2. There are 21 consonants. Select any 3 in 21C3 ways.
3. Arrange the 3 consonants in _C_C_C_ ways so that the 2 vowels are separated by them. So consonants are arranged in 3! ways.
4. We have 4 gaps above, so 2 vowels can be arranged in 4C2 * 2! ways.
5. Combine everything: 5C2 * (21C3 * 3!) * (4C2 * 2!)=957600(E).

bumpbot · Answer

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A 5-letter password must consist of different letters of the alphabet

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A 5-letter password must consist of different letters of the alphabet

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