<p>Consider the letters of the word MATHEMATICS. Possible number of words taking all letters at a time such that in each word both M's are together and both T's are together but both A's are not together is</p>
<p>\(\dfrac{11!}{2!2!2!} - \dfrac{10!}{2!2!}\)</p>
<p>\(7! \cdot {}^8C_2\)</p>
<p>\(\dfrac{6! \cdot 4!}{2! \cdot 2!} - \dfrac{10!}{2!2!2!}\)</p>
<p>\(\dfrac{9!}{2!2!2!}\)</p>
Step-by-Step Solution
Key Concept: Treat both M's as a single unit and both T's as a single unit, then use inclusion-exclusion principle: subtract cases where A's are together from cases where M's and T's are together.
<p><strong>Step 1: Identify the letters in MATHEMATICS</strong></p><p>Letters: M, A, T, H, E, M, A, T, I, C, S (11 letters total)</p><p>Frequency: M appears 2 times, A appears 2 times, T appears 2 times; H, E, I, C, S each appear 1 time</p><p><strong>Step 2: Treat M's as one unit and T's as one unit</strong></p><p>We now have: (MM), (TT), A, A, H, E, I, C, S = 9 objects with A appearing twice</p><p>Total arrangements of these 9 objects = 9!/2! = 181,440</p><p><strong>Step 3: Subtract cases where both A's are also together</strong></p><p>If M's together, T's together, AND A's together, we have: (MM), (TT), (AA), H, E, I, C, S = 8 distinct objects</p><p>Arrangements = 8! = 40,320</p><p><strong>Step 4: Apply inclusion-exclusion principle</strong></p><p>Words where M's together AND T's together BUT A's NOT together:</p><p>= (Arrangements with MM and TT together) − (Arrangements with MM, TT, and AA together)</p><p>= 181,440 − 40,320 = 141,120</p><p>∴ Answer: <strong>141,120</strong></p>
Correct Answer: A