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JEE Mathematics 2025 Question with Solution

All five letter words are made using all the letters A, B, C, D, E and arranged as in an English dictionary with serial numbers. Let the word at serial number nn be denoted by WnW_n. Let the probability P(Wn)P(W_n) of choosing the word WnW_n satisfy P(Wn)=2P(Wn−1)P(W_n) = 2P(W_{n-1}), n>1n > 1. If P(CDBEA)=2α2β−1P(CDBEA) = \frac{2^\alpha}{2^\beta - 1}, α,β∈N\alpha, \beta \in \mathbb{N}, then α+β\alpha + \beta is equal to :

Answer

Correct answer:183

Step-by-step solution

Standard Method

Given: All permutations of A, B, C, D, E are arranged in dictionary order, so the total number of words is 5!=1205! = 120. Also, P(Wn)=2P(Wn−1)P(W_n) = 2P(W_{n-1}) for n>1n > 1.

Find: The value of α+β\alpha + \beta if P(CDBEA)=2α2β−1P(CDBEA) = \frac{2^\alpha}{2^\beta - 1}.

First find the serial number of CDBEA.

Words beginning with A contribute

4!=244! = 24

Words beginning with B contribute

4!=244! = 24

Now the first letter is fixed as C.

For the second position, among the remaining letters A, B, D, E, the words beginning with CA and CB come before CD.

3!+3!=6+6=123! + 3! = 6 + 6 = 12

Now the first two letters are fixed as CD.

For the third position, among the remaining letters A, B, E, only A comes before B.

2!=22! = 2

Now the first three letters are fixed as CDB.

For the fourth position, among the remaining letters A, E, the word CDBAE comes before CDBEA.

1!=11! = 1

Hence, the number of words before CDBEA is

24+24+6+6+2+1=6324 + 24 + 6 + 6 + 2 + 1 = 63

Therefore, the rank of CDBEA is

6464

so CDBEA=W64CDBEA = W_{64}.

Let

P(W1)=xP(W_1) = x

Then the given relation gives

P(Wn)=2n−1xP(W_n) = 2^{n-1}x

for n=1,2,…,120n = 1,2,\dots,120.

Since total probability is 11,

∑n=1120P(Wn)=1\sum_{n=1}^{120} P(W_n) = 1

So,

x+2x+22x+⋯+2119x=1x + 2x + 2^2x + \cdots + 2^{119}x = 1

Using the sum of a geometric progression,

x(2120−12−1)=1x\left(\frac{2^{120}-1}{2-1}\right) = 1 x(2120−1)=1x(2^{120}-1) = 1 x=12120−1x = \frac{1}{2^{120}-1}

Compute the required probability

Now,

P(CDBEA)=P(W64)=263⋅P(W1)P(CDBEA) = P(W_{64}) = 2^{63} \cdot P(W_1)

Substituting P(W1)=12120−1P(W_1) = \frac{1}{2^{120}-1},

P(CDBEA)=263⋅12120−1=2632120−1P(CDBEA) = 2^{63} \cdot \frac{1}{2^{120}-1} = \frac{2^{63}}{2^{120}-1}

Comparing with

P(CDBEA)=2α2β−1P(CDBEA) = \frac{2^\alpha}{2^\beta - 1}

we get

α=63,β=120\alpha = 63, \quad \beta = 120

Hence,

α+β=63+120=183\alpha + \beta = 63 + 120 = 183

Therefore, the required answer is 183.

Common mistakes

  • A common mistake is to take the rank of CDBEA as 63 instead of 64. The count 6363 is the number of words preceding it, so the serial number is 63+1=6463 + 1 = 64.

  • Some students use the GP sum incorrectly by writing the last term as 2120x2^{120}x. Since P(Wn)=2n−1xP(W_n) = 2^{n-1}x, the terms run from xx to 2119x2^{119}x for 120120 words.

  • Another mistake is to forget that total probability over all 120120 words must be 11. Without imposing this normalization, P(W1)P(W_1) cannot be determined.

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