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JEE Mathematics 2023 Question with Solution

If the letters of the word MATHS are permuted and all possible words so formed are arranged as in a dictionary with serial number, then the serial number of the word THAMS is:

  • A

    102102

  • B

    103103

  • C

    101101

  • D

    104104

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: We need the rank of the word THAMS among all dictionary arrangements of the letters of MATHS.

Find: The serial number of THAMS.

Arrange the available letters in alphabetical order as A, H, M, S, T. Now count how many valid words come before THAMS.

For the first letter T, the smaller available letters are A, H, M, S. Each such choice gives

4!4!

words. So the number of words before those starting with T is

4×4!=4×24=96.4 \times 4! = 4 \times 24 = 96.

Now fix T and look at the second letter H among the remaining letters A, H, M, S. The only smaller available letter is A. For this choice, the remaining letters can be arranged in

3!=63! = 6

ways.

For the third letter A in THAMS, no smaller letter is available among the remaining letters. Similarly, for the fourth letter M and fifth letter S, no additional words occur before THAMS.

Hence, the total number of words before THAMS is

96+6=102.96 + 6 = 102.

Therefore, its rank is

102+1=103.102 + 1 = 103.

Therefore, the serial number of THAMS is 103103. The correct option is B.

Factorial Counting Detail

Given: The word is THAMS formed from the letters of MATHS.

Find: Its dictionary rank.

Using alphabetical order,

A<H<M<S<T.A < H < M < S < T.

So THAMS corresponds to the ordered positions T, H, A, M, S.

Count contribution place by place:

  1. Before T in the first place, there are 4 smaller letters. Contribution:
4⋅4!=96.4 \cdot 4! = 96.
  1. After fixing T, before H in the second place, there is 1 smaller letter, namely A. Contribution:
1⋅3!=6.1 \cdot 3! = 6.
  1. After fixing TH, before A in the third place, contribution is
0⋅2!=0.0 \cdot 2! = 0.
  1. After fixing THA, before M in the fourth place, contribution is
0⋅1!=0.0 \cdot 1! = 0.
  1. Last place contributes
0.0.

Thus,

96+6+0+0+0=102.96 + 6 + 0 + 0 + 0 = 102.

Adding 1 for the word itself,

Rank=102+1=103.\text{Rank} = 102 + 1 = 103.

Hence the correct option is B.

Common mistakes

  • A common mistake is to stop at 102102. That counts the number of words before THAMS, not the rank itself. Add 11 to get the serial number.

  • Another mistake is to use the letter order as they appear in MATHS instead of dictionary order. The counting must be based on A, H, M, S, T, not on the original arrangement.

  • Students may incorrectly count smaller letters before H after fixing T. At that stage only the remaining letters matter, and among A, H, M, S, only A is smaller than H.

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