All five-letter words are made using all the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $P(W_{1}) = x$. Since the total number of words is $5! = 120$,the sum of probabilities is $\sum_{i=1}^{120} P(W_{i}) = 1$. This is a geometric

Vedclass · Accepted Answer

$183$

Vedclass · Answer

(A) Let $P(W_{1}) = x$. Since the total number of words is $5! = 120$,the sum of probabilities is $\sum_{i=1}^{120} P(W_{i}) = 1$. This is a geometric progression: $x + 2x + 2^{2}x + \dots + 2^{119}x = 1$. $\frac{x(2^{120}-1)}{2-1} = 1 \Rightarrow x = \frac{1}{2^{120}-1}$. Now,we find the rank of the word $CDBEA$: Words starting with $A$: $4! = 24$. Words starting with $B$: $4! = 24$. Words starting with $CA$: $3! = 6$. Words starting with $CB$: $3! = 6$. Words starting with $CDA$: $2! = 2$. Words starting with $CDBAE$: $1$. Word $CDBEA$: $1$. Total rank $n = 24 + 24 + 6 + 6 + 2 + 1 + 1 = 64$. Thus,$P(W_{64}) = 2^{63} P(W_{1}) = \frac{2^{63}}{2^{120}-1}$. Comparing with $\frac{2^{\alpha}}{2^{\beta}-1}$,we get $\alpha = 63$ and $\beta = 120$. Therefore,$\alpha + \beta = 63 + 120 = 183$.

Similar Questions

The number of integers between $1$ and $10^{10}$ (inclusive) that contain the digit $1$ is:

The number of different words that can be formed from the letters of the word "$INTERMEDIATE$" such that two vowels never come together,is

The remainder obtained when $1! + 2! + 3! + \dots + 11!$ is divided by $12$ is

The sum of all possible numbers that can be formed by using the digits $2, 3, 5, 7$ without repetition of digits is

The total number of selections of the letters in the phrase "ned needs nineteen nets" is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module