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Permutations and Combinations question

2022 · 25 Jul · Shift 1 · Q37
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Permutations and Combinations question

2022 · 25 Jul · Shift 1 · Q37

JEE MainMathematicsPermutations and CombinationsNumerical+4 / −1
The letters of the word 'MANKIND' are written in all possible orders and arranged in serial order as in an English dictionary. Then the serial number of the word 'MANKIND' is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 1528

We need the dictionary rank of the word MANKIND among all permutations of its letters.

The letters are all distinct:

{A,D,I,K,M,N,N?}\{A, D, I, K, M, N, N?\}{A,D,I,K,M,N,N?}

Wait carefully — the word is MANKIND:

M,A,N,K,I,N,DM, A, N, K, I, N, DM,A,N,K,I,N,D

So the letters are:

A,D,I,K,M,N,NA, D, I, K, M, N, NA,D,I,K,M,N,N

There are two N's.

We now count how many valid words come before "MANKIND" in dictionary order.


1. First letter: M

Available letters: A,D,I,K,M,N,NA, D, I, K, M, N, NA,D,I,K,M,N,N

Letters smaller than MMM are:

A,D,I,KA, D, I, KA,D,I,K

If the first letter is any one of these, the remaining 666 letters can be arranged in

6!2!=360\frac{6!}{2!} = 3602!6!​=360

ways (because two N's are identical).

So total words before those starting with MMM:

4×360=14404 \times 360 = 14404×360=1440


2. Fix first letter M, second letter: A

Now remaining letters: D,I,K,N,ND, I, K, N, ND,I,K,N,N

The second letter in the word is already the smallest among available letters, so no word is contributed here.

Count added:

000


3. Fix MA, third letter: N

Remaining letters: D,I,K,ND, I, K, ND,I,K,N

Letters smaller than NNN are:

D,I,KD, I, KD,I,K

For each such choice, remaining 444 letters can be arranged in

4!=244! = 244!=24

ways.

So contribution:

3×24=723 \times 24 = 723×24=72

Running total:

1440+72=15121440 + 72 = 15121440+72=1512


4. Fix MAN, fourth letter: K

Remaining letters: D,I,ND, I, ND,I,N

Letters smaller than KKK are:

D,ID, ID,I

For each choice, remaining 333 letters can be arranged in

3!=63! = 63!=6

ways.

So contribution:

2×6=122 \times 6 = 122×6=12

Running total:

1512+12=15241512 + 12 = 15241512+12=1524


5. Fix MANK, fifth letter: I

Remaining letters: D,ND, ND,N

Letters smaller than III are:

DDD

Then remaining 222 letters can be arranged in

2!=22! = 22!=2

ways.

Contribution:

1×2=21 \times 2 = 21×2=2

Running total:

1524+2=15261524 + 2 = 15261524+2=1526


6. Fix MANKI, sixth letter: N

Remaining letters: D,ND, ND,N

Letters smaller than NNN are:

DDD

Then the remaining letters are NNN, so only

1!=11! = 11!=1

arrangement.

Contribution:

111

Running total:

1526+1=15271526 + 1 = 15271526+1=1527


7. Fix MANKIN, last letter is D

No further contribution.

Thus, number of words before MANKIND is

152715271527

Hence its serial number is

1527+1=15281527 + 1 = 15281527+1=1528


Final Answer

1528\boxed{1528}1528​


Comparison with stored answer

Stored correct answer: 149214921492

Our derived answer is 152815281528, so they do not match.

The likely reason the stored answer is incorrect is that it seems to treat the letters as all distinct, but the word MANKIND contains two identical N's, so permutations must be counted using division by 2!2!2! where appropriate.

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