Question

If the number of all possible permutations of the letters of the word MATHEMATICS in which the repeated letters are not together is 90(X), then X =

Answer 1

30800

Answer 2

The word is MATHEMATICS. Let's count the frequency of each letter:

M: 2 times A: 2 times T: 2 times H: 1 time E: 1 time I: 1 time C: 1 time S: 1 time

Total number of letters = 11.

1. Calculate the total number of permutations ($N_{total}$):

The formula for permutations with repetitions is $\frac{n!}{n_1! n_2! ... n_k!}$. Here, n=11, n_M=2, n_A=2, n_T=2.

$N_{total} = \frac{11!}{2! \times 2! \times 2!} = \frac{39,916,800}{8} = 4,989,600$.

2. Calculate the number of permutations where repeated letters are together:

We use the Principle of Inclusion-Exclusion. Let $P_M$ be the property that the two M's are together (MM). Let $P_A$ be the property that the two A's are together (AA). Let $P_T$ be the property that the two T's are together (TT). We need to find the number of permutations where at least one of these properties holds, i.e., $N(P_M \cup P_A \cup P_T)$.

$N(P_M \cup P_A \cup P_T) = \sum N(P_M) - \sum N(P_M \cap P_A) + N(P_M \cap P_A \cap P_T)$.

• Sum of individual properties ($\sum N(P_M)$):

  To find $N(P_M)$, treat 'MM' as a single block. We are now permuting 10 items: (MM), A, T, H, E, A, T, I, C, S. The repeated letters among these are A (2 times) and T (2 times).

  $N(P_M) = \frac{10!}{2! \times 2!} = \frac{3,628,800}{4} = 907,200$.

  By symmetry, $N(P_A) = 907,200$ and $N(P_T) = 907,200$.

  $\sum N(P_M) = 3 \times 907,200 = 2,721,600$.

• Sum of properties taken two at a time ($\sum N(P_M \cap P_A)$):

  To find $N(P_M \cap P_A)$, treat 'MM' as one block and 'AA' as another block. We are now permuting 9 items: (MM), (AA), T, H, E, T, I, C, S. The repeated letter among these is T (2 times).

  $N(P_M \cap P_A) = \frac{9!}{2!} = \frac{362,880}{2} = 181,440$.

  By symmetry, $N(P_M \cap P_T) = 181,440$ and $N(P_A \cap P_T) = 181,440$.

  $\sum N(P_M \cap P_A) = 3 \times 181,440 = 544,320$.

• Property taken three at a time ($N(P_M \cap P_A \cap P_T)$):

  To find $N(P_M \cap P_A \cap P_T)$, treat 'MM', 'AA', and 'TT' as single blocks. We are now permuting 8 items: (MM), (AA), (TT), H, E, I, C, S. All these 8 items are distinct.

  $N(P_M \cap P_A \cap P_T) = 8! = 40,320$.

• Calculate $N(P_M \cup P_A \cup P_T)$:

  $N(P_M \cup P_A \cup P_T) = 2,721,600 - 544,320 + 40,320 = 2,177,280 + 40,320 = 2,217,600$.

  This is the number of permutations where at least one pair of repeated letters is together.

3. Calculate the number of permutations where repeated letters are not together:

This is given by $N_{total} - N(P_M \cup P_A \cup P_T)$.

$N_{not together} = 4,989,600 - 2,217,600 = 2,772,000$.

4. Solve for X:

The problem states that the number of such permutations is 90(X).

So, $90(X) = 2,772,000$.

$X = \frac{2,772,000}{90} = \frac{277,200}{9} = 30,800$.

The final answer is $\boxed{30800}$.

Explanation of the solution:

1. Calculate total permutations of the word MATHEMATICS: $\frac{11!}{2!2!2!} = 4,989,600$.

2. Use the Principle of Inclusion-Exclusion to find permutations where at least one pair of repeated letters (MM, AA, TT) is together.

   • Sum of permutations with one pair together (e.g., MM as a block): $3 \times \frac{10!}{2!2!} = 3 \times 907,200 = 2,721,600$.

   • Sum of permutations with two pairs together (e.g., MM and AA as blocks): $3 \times \frac{9!}{2!} = 3 \times 181,440 = 544,320$.

   • Permutations with all three pairs together (MM, AA, TT as blocks): $8! = 40,320$.

   • Number of permutations with at least one pair together = $2,721,600 - 544,320 + 40,320 = 2,217,600$.

3. Subtract this from the total permutations to find permutations where no repeated letters are together: $4,989,600 - 2,217,600 = 2,772,000$.

4. Equate this to $90(X)$ and solve for X: $2,772,000 = 90(X) \implies X = 30,800$.