To solve the problem, we need to find the number of arrangements of the letters in the word "MATHEMATICS" such that the arrangement contains the letters "MATH" as a block. Then, we will determine the number of divisors of that result.
### Step 1: Identify the letters in "MATHEMATICS"
The word "MATHEMATICS" consists of the following letters:
- M: 2
- A: 2
- T: 2
- H: 1
- E: 1
- I: 1
- C: 1
- S: 1
### Step 2: Treat "MATH" as a single unit
Since we want the arrangement to contain "MATH," we can treat "MATH" as a single unit or block. Let's denote this block as "G".
Now, the letters we have are:
- G (which represents "MATH")
- E
- M (remaining 1)
- A (remaining 1)
- T (remaining 1)
- I
- C
- S
This gives us a total of 8 units to arrange: G, E, M, A, T, I, C, S.
### Step 3: Calculate the arrangements of the blocks
Now, we need to calculate the number of arrangements of these 8 units. The formula for the arrangements of n items where some items are identical is given by:
\[
\text{Number of arrangements} = \frac{n!}{p_1! \times p_2! \times \ldots}
\]
In our case, we have:
- Total units = 8 (G, E, M, A, T, I, C, S)
- The letters M and A are repeated.
So the number of arrangements is:
\[
\text{Number of arrangements} = \frac{8!}{1! \times 1! \times 1! \times 1! \times 1! \times 2! \times 2!} = \frac{8!}{2! \times 2!}
\]
Calculating \(8!\):
\[
8! = 40320
\]
Calculating \(2!\):
\[
2! = 2
\]
Thus, we have:
\[
\text{Number of arrangements} = \frac{40320}{2 \times 2} = \frac{40320}{4} = 10080
\]
### Step 4: Find the number of divisors of 10080
Next, we need to find the prime factorization of 10080.
To factor 10080, we can divide it by prime numbers:
\[
10080 = 10080 \div 2 = 5040
\]
\[
5040 \div 2 = 2520
\]
\[
2520 \div 2 = 1260
\]
\[
1260 \div 2 = 630
\]
\[
630 \div 2 = 315
\]
\[
315 \div 3 = 105
\]
\[
105 \div 3 = 35
\]
\[
35 \div 5 = 7
\]
So, the prime factorization of 10080 is:
\[
10080 = 2^5 \times 3^2 \times 5^1 \times 7^1
\]
### Step 5: Calculate the number of divisors
The formula for finding the number of divisors \(d(n)\) from the prime factorization \(p_1^{a_1} \times p_2^{a_2} \times \ldots\) is:
\[
d(n) = (a_1 + 1)(a_2 + 1)(a_3 + 1) \ldots
\]
For our factorization \(2^5 \times 3^2 \times 5^1 \times 7^1\):
- \(a_1 = 5\) (for 2)
- \(a_2 = 2\) (for 3)
- \(a_3 = 1\) (for 5)
- \(a_4 = 1\) (for 7)
Thus, the number of divisors is:
\[
d(10080) = (5 + 1)(2 + 1)(1 + 1)(1 + 1) = 6 \times 3 \times 2 \times 2
\]
Calculating this gives:
\[
d(10080) = 6 \times 3 = 18
\]
\[
18 \times 2 = 36
\]
\[
36 \times 2 = 72
\]
### Final Answer
The number of divisors of \(a\) (which is 10080) is **72**.
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