Solution:
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
$$ = \left( {{}^4{C_1} \times {}^5{C_4}} \right) + \left( {{}^4{C_2} \times {}^5{C_3}} \right)$$      $$ + \left( {{}^4{C_3} \times {}^5{C_2}} \right)$$   $$ + \left( {{}^4{C_6} \times {}^5{C_1}} \right)$$
$$ = \left( {{}^4{C_1} \times {}^5{C_1}} \right) + \left( {{}^4{C_2} \times {}^5{C_2}} \right)$$      $$ + \left( {{}^4{C_1} \times {}^5{C_2}} \right)$$   $$ + \left( {{}^4{C_4} \times {}^5{C_1}} \right)$$
$$ = \left( {4 \times 5} \right) + \left( {\frac{{4 \times 3}}{{2 \times 1}} \times \frac{{5 \times 4}}{{2 \times 1}}} \right)$$      $$ + \left( {4 \times \frac{{5 \times 4}}{{2 \times 1}}} \right)$$   $$ + \left( {1 \times 5} \right)$$
$$ = \left( {20 + 60 + 40 + 5} \right)$$
$$ = 125$$

Solution:
Number of ways of choosing 6 from 10 = ¹⁰C₆ = 210 Number of ways of attempting more than 4 from a group, = 2 × ⁵C₅ × ⁵C₁
= 10 Required number of ways
= 210 - 10
= 200

Solution:
If the word started with the letter A then the remaining 5 positions can be filled in 5! Ways. If it started with c then the remaining 5 positions can be filled in 5! Ways. Similarly if it started with H, I, N the remaining 5 positions can be filled in 5! Ways. If it started with S then the remaining position can be filled with A, C, H, I, N in alphabetical order as on dictionary. The required word SACHIN can be obtained after the 5 × 5! = 600 Ways. i.e. SACHIN is the 601^th word.

Solution:
First place can be filled in 4 ways. The subsequent places can be filled in 3 ways each. Hence, the number of ways
= 4 × 3 × 3 × 3 × 3 × 3
= 4 × 3⁵

Solution:
Let total number of persons present in the party be x, Then, $$\eqalign{ \frac{{x \times \left( {x - 1} \right)}}{2} = 105 \cr x = 15 \cr} $$

Solution:
The given word contains 6 letter out of which R is taken 2 times, U is taken to 2 times and other letters are all different.
∴ Required number of ways
$$\eqalign{ = \frac{{6!}}{{2! \times 2!}} \cr = \frac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 2}} \cr = 180 \cr} $$

Solution:
The given words contains 6 letters of which E is taken 2 times.
∴ Required number of ways
$$\eqalign{ = \frac{{6!}}{{2!}} \cr = \frac{{6 \times 5 \times 4 \times 3 \times 2!}}{{2!}} \cr = 360 \cr} $$