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