Answer: Option C. -> 10 : 1
Initially we look at the general case of the seats not numbered.
The total number of cases of arranging 8 men and 2 women, so that women are together,
⇒ 8! ×2!
The number of cases where in the women are not together,
⇒ 9! - (8! × 2!) = Q
Now, when the seats are numbered, it can be considered to a linear arrangement and the number of ways of arranging the group such that no two women are together is,
⇒ 10! - (9! × 2!)
But the arrangements where in the women occupy the first and the tenth chairs are not favorable as when the chairs which are assumed to be arranged in a row are arranged in a circle, the two women would be sitting next to each other.
The number of ways the women can occupy the first and the tenth position,
= 8! × 2!
The value of P = 10! - (9! × 2!) - (8! × 2!)
Thus P : Q = 10 : 1

Answer: Option B. -> 2520
The given word contains 7 letters which D is taken 2 times.
∴ Required number of ways
$$\eqalign{
& = \frac{{7!}}{{2!}} \cr
& = \frac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}} \cr
& = 2520 \cr} $$

Answer: Option A. -> 840
The given word contains 7 letters of which D is taken 3 times.
∴ Required number of ways
$$\eqalign{
& = \frac{{7!}}{{3!}} \cr
& = \frac{{7 \times 6 \times 5 \times 4 \times 3!}}{{3!}} \cr
& = \left( {7 \times 6 \times 5 \times 4} \right) \cr
& = 840 \cr} $$

Answer: Option D. -> 576
The given word contains 7 different letters.
Keeping the vowels (AUIO) together, we take them as 1 letter.
Then, we have to arrange the letters CTN (AUIO).
Now, 4 letters can be arranged in 4! = 24 ways.
The vowels (AUIO) can be arranged among themselves in 4! = 24 ways.
∴ Required number of ways = (24 × 24) = 576

Answer: Option C. -> 931
Required number of ways
$$ = \left( {{}^6{C_1} \times {}^8{C_3}} \right) + \left( {{}^6{C_2} \times {}^8{C_2}} \right)$$     $$ + \left( {{}^6{C_3} \times {}^8{C_1}} \right)$$   $$ + \left( {{}^6{C_4} \times {}^8{C_0}} \right)$$
$$ = \left\{ {6 \times \frac{{8 \times 7 \times 6}}{{3 \times 2 \times 1}}} \right\} + $$    $$\left( {\frac{{6 \times 5}}{{2 \times 1}} \times \frac{{8 \times 7}}{{2 \times 1}}} \right)$$   $$ + \left( {\frac{{6 \times 5 \times 4}}{{3 \times 2 \times 1}} \times 8} \right)$$    $$ + \left( {{}^6{C_2} \times 1} \right)$$
$$ = \left\{ {6 \times \frac{{8 \times 7 \times 6}}{{3 \times 2 \times 1}}} \right\}$$    $$ +\, 420\, + $$  $$\left( {\frac{{6 \times 5 \times 4}}{6} \times 8} \right)$$   $$ + \left( {\frac{{6 \times 5}}{{2 \times 1}} \times 1} \right)$$
$$ = \left( {336 + 420 + 160 + 15} \right)$$
$$ = 931$$

Answer: Option B. -> 576
There are 7 letters in the given word, out of which there are 3 vowels and 4 consonants.
Let us mark the positions to be filled up as follows:
$$\left( {\mathop {}\limits^1 } \right)\left( {\mathop {}\limits^2 } \right)\left( {\mathop {}\limits^3 } \right)\left( {\mathop {}\limits^4 } \right)\left( {\mathop {}\limits^5 } \right)\left( {\mathop {}\limits^6 } \right)\left( {\mathop {}\limits^7 } \right)$$
Now, 3 vowels can placed at any of the three places out of four marked 1, 3, 5, 7
Number of ways of arranging the vowels
$$\eqalign{
& = {}^4{P_3} \cr
& = \left( {4 \times 3 \times 2} \right) \cr
& = 24 \cr} $$
4 consonants at the remaining 4 positions may be arranged in $${}^4{P_4} = 4! = $$  
24 ways
Required number of ways = (24 × 24) = 576

Answer: Option A. -> 360
There are six letters in the given word MOMENT and letter 'M' has come twice.
∴ Required number of ways
$$\eqalign{
& = \frac{{6!}}{{2!}} \cr
& = \frac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}} \cr
& = 360 \cr} $$

Answer: Option C. -> 125
The committee should have
(1 man, 4 women) or (2 men, 3 women) or (3 men, 2 women) or ( 4 men, 1 woman)
Required number of ways
$$ = \left( {{}^4{C_1} \times {}^5{C_4}} \right) + \left( {{}^4{C_2} \times {}^5{C_3}} \right)$$      $$ + \left( {{}^4{C_3} \times {}^5{C_2}} \right)$$   $$ + \left( {{}^4{C_6} \times {}^5{C_1}} \right)$$
$$ = \left( {{}^4{C_1} \times {}^5{C_1}} \right) + \left( {{}^4{C_2} \times {}^5{C_2}} \right)$$      $$ + \left( {{}^4{C_1} \times {}^5{C_2}} \right)$$   $$ + \left( {{}^4{C_4} \times {}^5{C_1}} \right)$$
$$ = \left( {4 \times 5} \right) + \left( {\frac{{4 \times 3}}{{2 \times 1}} \times \frac{{5 \times 4}}{{2 \times 1}}} \right)$$      $$ + \left( {4 \times \frac{{5 \times 4}}{{2 \times 1}}} \right)$$   $$ + \left( {1 \times 5} \right)$$
$$ = \left( {20 + 60 + 40 + 5} \right)$$
$$ = 125$$

Answer: Option B. -> 120960
Keeping the vowels (AEIA) together, we have MTHMTCS (AEAI).
Now, we have to arrange 8 letters, out of which we have 2M, 2T and the rest are all different.
Number of ways of arranging these letters
$$\eqalign{
& = \frac{{8!}}{{2!.2!}} \cr
& = \frac{{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 2 \times 1}} \cr
& = 10080 \cr} $$
Now, (AEAI) has 4 letters, out of which we have 2A, 1E and 1I.
Number of ways of arranging these letters
$$\eqalign{
& = \frac{{4!}}{{2!}} \cr
& = \frac{{4 \times 3 \times 2 \times 1}}{2} \cr
& = 12 \cr} $$
∴ Required number of ways = (10080 × 12) = 120960

Answer: Option E. -> 4320
The given word contains 8 different letters.
We keep the vowels (OAE) together and treat them as 1 letter.
Thus, we have to arrange the 6 letters SFTWR(OAE)
These can be arranged in 6! = 720 ways
The vowels (OAE) can be arranged among themselves in 3! = 6 ways.
∴ Required number of ways = (720 × 6) = 4320