Permutations and Combinations Part 2

We are given that the probability of choosing Wn is:
‌P(Wn)=2P(Wn−1)‌ for ‌n>1
‌⇒P(W1)=p,P(W2)=2p,P(W3)=4p,...,P(Wn)=2n−1p

To find the value of p, we use:

120

∑

n−1

P(Wn)=1‌ (total probability) ‌
So, p(1+2+22+...+2119)=1
p(2120−1)=1⇒p=‌

1

2120−1

Thus, P(Wn)=‌

2n−1

2120−1

....(i)

Since the first letter of CDBEA is C
Words starting with A:4!=24
Words starting with B:4!=24
‌CA−−−:3!=6
‌CB_−_:3!=6
‌CDA−−:2!=2
‌CDBA−:1!=1
Total before CDBEA=63
Position of CDBEA =64‌th ‌
Putting in (i)
P(CDBEA)=P(W64)=‌

263

2120−1

Compare with the given form:
P(CDBEA)=‌

2α

2β−1

Compare with the given form:
P(CDBEA)=‌

2α

2β−1

So, α=63,β=120
α+β=63+120=183