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Permutations and Combinations question

2025 · 3 Apr · Shift 1 · Q46
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Permutations and Combinations question

2025 · 3 Apr · Shift 1 · Q46

JEE MainMathematicsPermutations and CombinationsNumerical+4 / −1
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 nnn be denoted by Wn\mathrm{W}_{\mathrm{n}}Wn​. Let the probability P(Wn)\mathrm{P}\left(\mathrm{W}_{\mathrm{n}}\right)P(Wn​) of choosing the word Wn\mathrm{W}_{\mathrm{n}}Wn​ satisfy P(Wn)=2P(Wn−1),n>1\mathrm{P}\left(\mathrm{W}_{\mathrm{n}}\right)=2 \mathrm{P}\left(\mathrm{W}_{\mathrm{n}-1}\right), \mathrm{n}\gt 1P(Wn​)=2P(Wn−1​),n>1. If P(CDBEA)=2α2β−1,α,β∈N\mathrm{P}(\mathrm{CDBEA})=\frac{2^\alpha}{2^\beta-1}, \alpha, \beta \in \mathbb{N}P(CDBEA)=2β−12α​,α,β∈N, then α+β\alpha+\betaα+β is equal to : ‾\underline{\hspace{2cm}}​
Numerical answer
View written solutionFree

Correct answer: 183

  1. Total number of words

Using all the letters A,B,C,D,EA,B,C,D,EA,B,C,D,E, the total number of 5-letter words is 5!=120.5! = 120.5!=120.

They are arranged in dictionary order and denoted by W1,W2,…,W120W_1, W_2, \dots, W_{120}W1​,W2​,…,W120​.


  1. Probability pattern

Given P(Wn)=2P(Wn−1),n>1.P(W_n)=2P(W_{n-1}), \quad n>1.P(Wn​)=2P(Wn−1​),n>1. So the probabilities form a geometric progression: P(Wn)=2n−1P(W1).P(W_n)=2^{n-1}P(W_1).P(Wn​)=2n−1P(W1​).

Since total probability is 1, ∑n=1120P(Wn)=1.\sum_{n=1}^{120} P(W_n)=1.∑n=1120​P(Wn​)=1. Thus, P(W1)(1+2+22+⋯+2119)=1.P(W_1)(1+2+2^2+\cdots+2^{119})=1.P(W1​)(1+2+22+⋯+2119)=1. Using geometric sum, P(W1)(2120−1)=1,P(W_1)(2^{120}-1)=1,P(W1​)(2120−1)=1, so P(W1)=12120−1.P(W_1)=\frac{1}{2^{120}-1}.P(W1​)=2120−11​.

Hence, P(Wn)=2n−12120−1.P(W_n)=\frac{2^{n-1}}{2^{120}-1}.P(Wn​)=2120−12n−1​.


  1. Find the serial number of CDBEA\mathrm{CDBEA}CDBEA

We now find the rank of the word CDBEA\mathrm{CDBEA}CDBEA in dictionary order.

Position 1: C

Words starting with AAA come first, then words starting with BBB. For each fixed first letter, remaining letters can be arranged in 4!=244! = 244!=24 ways.

So words before those starting with CCC: 2⋅24=48.2 \cdot 24 = 48.2⋅24=48.


Position 2: D

Now consider words starting with CCC. Remaining letters are A,B,D,EA,B,D,EA,B,D,E. Among these, letters smaller than DDD are A,BA,BA,B. For each such choice at second position, remaining 3 letters can be arranged in 3!=63! = 63!=6 ways.

So additional words before CDCDCD: 2⋅6=12.2 \cdot 6 = 12.2⋅6=12.

Running total: 48+12=60.48+12=60.48+12=60.


Position 3: B

Now prefix is CDCDCD. Remaining letters are A,B,EA,B,EA,B,E. Letters smaller than BBB is only AAA. For this choice, remaining 2 letters can be arranged in 2!=22! = 22!=2 ways.

Additional words: 1⋅2=2.1\cdot 2=2.1⋅2=2.

Running total: 60+2=62.60+2=62.60+2=62.


Position 4: E

Now prefix is CDBCDBCDB. Remaining letters are A,EA,EA,E. Letter smaller than EEE is AAA. Then remaining 1 letter can be arranged in 1!=11! = 11!=1 way.

Additional words: 1.1.1.

Running total: 62+1=63.62+1=63.62+1=63.


Position 5: A

Now only one word remains with prefix CDBECDBECDBE, namely CDBEACDBEACDBEA itself. So its rank is 63+1=64.63+1=64.63+1=64.

Therefore, CDBEA=W64.\mathrm{CDBEA}=W_{64}.CDBEA=W64​.


  1. Compute its probability

Using P(Wn)=2n−12120−1,P(W_n)=\frac{2^{n-1}}{2^{120}-1},P(Wn​)=2120−12n−1​, we get P(CDBEA)=P(W64)=2632120−1.P(\mathrm{CDBEA})=P(W_{64})=\frac{2^{63}}{2^{120}-1}.P(CDBEA)=P(W64​)=2120−1263​.

This matches the form 2α2β−1.\frac{2^\alpha}{2^\beta-1}.2β−12α​. Thus, α=63,β=120.\alpha=63, \quad \beta=120.α=63,β=120. So, α+β=63+120=183.\alpha+\beta=63+120=183.α+β=63+120=183.


  1. Comparison with stored answer

Derived answer is 183,183,183, which matches the stored correct answer.

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