What is the probability that out of the combinations that can be made

Question

What is the probability that out of the combinations that can be made using all the letters of the word EXCESS, Jerome will randomly pick a combination in which the first letter is a vowel and the last letter is a consonant?

A. 96/320
B. 24/180
C. 33/100
D. 48/180
E. 96/180

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A. 96/320
B. 24/180
C. 33/100
D. 48/180
E. 96/180

Are You Up For the Challenge: 700 Level Questions

Archit3110 · Answer

total ways for EXCESS ; 6!/2!*2! = 180
and first letter is a vowel and the last letter is a consonant 
2*1 * 1*2*3*4 = 48
IMO D: 48/180

harivars · Answer

can u plz explain how do u got 48

Archit3110 · Answer

harivars

EXCESS has 6 words 
Vowel = 2
Consonant=4
condition to be followed the first letter is a vowel and the last letter is a consonant
2*1*1*2*3*4 = 48

What is the probability that out of the combinations that can be made using all the letters of the word EXCESS, Jerome will randomly pick a combination in which the first letter is a vowel and the last letter is a consonant?

A. 96/320
B. 24/180
C. 33/100
D. 48/180
E. 96/180

Are You Up For the Challenge: 700 Level Questions

satya2029 · Answer

EXCESS
2-E
2-S
1-X
1-C
First place vowel and last place consonant
first E at first place only one way
for second E remaining places except the last one- so 4 places
last place consonant-4C1/!2 because of 2S
total ways=1*4*12
=48
Required ways=48/(!6/!2*!2)
D:)

stne · Answer

Why do you have 1 in the second place? 
after we have chosen the first and last we are left with 4 characters. So why is it not 2*4*3*2*1*4 ?

Kushchokhani · Answer

I solved this ques making 2 cases-
Case 1: If S is the last letter-
then (2!/2!)*1*4! ways= 24 ways

in details- (2!/2!) is for selecting vowel; 1 is for selecting last letter S; 4! is for arranging middle letters, which are all different

Case 2: If S is not the last letter-
then (2!/2!)*2!*(4!/2!) ways= 24 ways

in details- (2!/2!) is for selecting vowel; 2! is for selecting last letter from either X or C; (4!/2!) is for arranging middle letters, among which S is repeating

Total favorable ways=48

adipisci · Answer

Your point is correct and no one addressed it properly. It will be 48 because of the following explanation -

For first vowel - 2 options
For last consonant - 4 options
For 4 places in between - 4! options

But there are 2 E and 2 S. So the product will be divided by 2! . 2!.

So,
(2*4*3*2*4) / 2!.2! = 48.

Hope it makes sense.

hemi11984 · Answer

There are two cases:
(I) E_ _ _ _ S for which E can be at first place in 1 way and S can be selected at last place in 1 way. Rest are EXCS and four places so total arrangement is 4! = 24.
(II) E_ _ _ _ (X/C) in this case E can be selected for 1 place in 1 way. X or C can be selected at the last place in 2 ways and rest 4 letters ESS and X or C (if X is at last place then C or vice versa) can be arranged in 4!/2!, therefore total number of combination in this case is 2 x (4!/2!) = 24.
Therefore total number of combination in which the first letter is a vowel and the last letter is a consonant is 24 + 24 = 48.
Total number of combination for the letters EXCESS is 6! /2!*2! = 180.
Hence required probability is 48/180. Ans (D)

What is the probability that out of the combinations that can be made

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What is the probability that out of the combinations that can be made

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