What is the probability that an arrangement of a b c d e f h

What is the probability that an arrangement of a, b, c, d, e, f has

(a) a and b side-by-side?

(b) a occurring somewhere before b?

Solution

the total number of arrangements os the letters a,b,c,d,e,f are 6!

which give 720 arrangemnts thew letters can be arranged in 6! ways

now for a, b to side by side consider both a and b to be one unit

so the total no of units are (a,b),c,d,e,f which are 5 \\

so the total arrangements are 5! which is 120 but the letters a and b can interchange themselves in 2 two way s as (a,b) and (b,a)

so the probability that a, b appear side by side is P= 5!*2/6!= (120*2)/720 = 0.333

the second case of a occuring somewhere before b is excluding the case where a,b interchnage themseves and considering a,b as single non interchangable unit which gives

5 units in total so 5! arrangements

probabilty that a occuring before b is P= 51/6!=120/720 = 0.166