If the letters a, A, b, B, c, and C are arranged at random in a row

Question

If the letters a, A, b, B, c, and C are arranged at random in a row, what is the probability that the lower case letter appear in increasing alphabetical order?

A. 1/2
B. 1/3
C. 1/4
D. 1/6
E. 1/12

Bunuel · Accepted Answer

Three letters a, b, and c can be arranged in 3!=6 different ways and only one of them will be in increasing alphabetical order, namely: a-b-c, so P=1/6.

sjgudapa · Answer

Bunnel,

Need your help in understanding this.

If they are asking for the probability of arranging the lower case abc in alphabetical order

I am assuming that first a,b,c and then A , B, C can we be arranged in 3! possible ways.

Desired outcomes = 1 X 3!

Total possible outcomes = 6! as all the 6 letters can be arranged in 6! ways

hence prob = 3! / 6! = 1/120.

Please comment on where I am going wrong?

Thanks!

imania · Answer

hey Bunuel
what's wrong with my approach?
A,a,b,B,c,C
desired mode:abc
total :6!
p=4!/6!

Bunuel · Answer

The problem with both solutions above is that favorable outcomes are much more, namely 120.

{***}{a}{b}{c} -  4*3!=24  (capital letters are together);
{**}{a}{*}{b}{c} -  C^2_3*2*4*3=72  (2 capital letters are together);
{*}{a}{*}{b}{*}{c} -  C^3_4*3!=24  (capital letters are separated);

24+72+24=120  -->  P=\frac{120}{6!}=\frac{1}{6} .

But this is a long way of solving. Consider one particular arrangement: A*B*C*, lower case letters for *. We can arrange lower case letters instead of * in 3!=6 ways but only one will be in alphabetical order AaBbCc, so 1 out of 6. For other such cases also only one out of 6 will be in alphabetical order (ABCabc, ....), so P=1/6.

Basically we can ignore capital letters for this problem and say: 3 letters a, b, and c can be arranged in 3!=6 different ways and only one of them will be in increasing alphabetical order, namely: a-b-c, so P=1/6.

5va · Answer

Hi Bunuel,

Even i also think that answer should be 4!/6!

shrouded1 · Answer

Think of these arrangements as the following :

___ ___ ___ ____

Where   act as placeholders for the smaller case alphabets. And for each arrangement of small case letters, the ___ are used as placeholders for one of more upper case letters.

For Eg. The arrangement of small case letters __b__a__c___ can then produce several final arrangements such as ABCbac or bACaBc etc etc

Now all I am going to say is that for each arrangement of type ___b___a___c___ there are an equal number of final arrangements possible. As you just change the small case letters used in your placeholders keeping the upper case ones constant.

How many types of arrangement exist ? The number of ways to place small case letters in place holders which is 3!
How many of these have a,b,c in order ? Just 1 : __a__b__c__

So probability = 1/6

prashantbacchewar · Answer

I am totally confused

Bunuel can you pls simplyfy the answer for me

Thanks in advance..

Bunuel · Answer

Can you please tell what exactly didn't you understand?

Simplifying: you can ignore capital letters for this problem. So, 3 letters a, b, and c can be arranged in 3!=6 different ways and only one of them will be in increasing alphabetical order, namely: a-b-c, so P=1/6.

Souvik7 · Answer

1. Total possible cases = arrangement of 6 letters = 6! 
2. Sample cases ; 
There are 6 spots for those letters , 
- - - - - - we will chose 3 spots for abc in alphabetical order in 6C3 ways = 20 ways

& remaining ABC can arrange themselves in 3! Ways.

So total = 6C3 * 3! =120

Req ans = 120/6! = 1/6

Ans =

Posted from my mobile device

GSV · Answer

Hi Bunuel,

Can I know why we are not considering A, B, C. 
And another question based on below replies, why can't abc be together. Nowhere in the question it is mentioned that a, b, c should be seperate.

Your reply would be of a lot of help.

Bunuel · Answer

We do  not  need to consider A, B, C because the uppercase letters do  not  affect the  relative order  of a, b, c. No matter where A, B, C land, the lowercase letters can appear in exactly 6 possible orders. Only 1 of those 6 is a-b-c.

About “why can’t abc be together?” They  can  be together. It does not matter. Whether they are adjacent or separated, the probability that their  order  is a-b-c is still 1 out of the 6 possible orders.

OneLazyBum · Answer

How about we do (6c3)/(6P3)
In numerator 3 letters can be arranged at 6 positions, in 6c3 ways and we don't multiply it with 3! because only one order will be correct
but for total number of ways we do (6P3) because order matters here

If the letters a, A, b, B, c, and C are arranged at random in a row

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If the letters a, A, b, B, c, and C are arranged at random in a row

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