Solution:
Required number of ways
$$ = \left( {{}^7{C_3} \times {}^6{C_2}} \right) + $$   $$\left( {{}^7{C_4} \times {}^6{C_1}} \right) + $$   $$\left( {{}^7{C_5} \times {}^6{C_0}} \right)$$
$$ = \left\{ {\frac{{7 \times 6 \times 5}}{{3!}} \times \frac{{6 \times 5}}{{2!}}} \right\}$$     $$ + \left( {{}^7{C_3} \times {}^6{C_1}} \right)$$   $$ + \left( {{}^7{C_2} \times 1} \right)$$
$$ = \left\{ {\frac{{7 \times 6 \times 5}}{6} \times \frac{{6 \times 5}}{{2 \times 1}}} \right\}$$     $$ + \left( {\frac{{7 \times 6 \times 5}}{{3 \times 2 \times 1}} \times 6} \right)$$    $$ + \left( {\frac{{7 \times 6}}{{2 \times 1}} \times 1} \right)$$
$$\eqalign{ = \left( {525 + 210 + 21} \right) \cr = 756 \cr} $$

Solution:
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

Solution:
The four person who wish to sit facing forward can be seated in: ⁵P₄ ways and 3 who wish to sit facing towards the rear can be seated in: ⁵P₃ ways and the remaining 3 can be seated in the remaining 3 seats in ³P₃ ways.
Total number of ways
= ⁵P₄ × ⁵P₃ × ³P₃
= 43200

Solution:
Three possibilities: (1W+6M), (2W+5M), (3W+4M) = (⁶C₁ × ⁹C₆) + (⁶C₂ × ⁹C₅) + (⁶C₃ × ⁹C₄)
= 4914

Solution:
ⁿP₃ = n × (n–1) × (n–2) ⁽ⁿ⁺¹⁾P₃ = (n+1) × n × (n–1) Now, 5 × n × (n–1) × (n–2) = 4 × (n+1) × n × (n–1) Or, 5(n−2) = 4(n+1) Or, 5n − 10 = 4n + 4 Or, 5n − 4n = 4 + 10 Hence, n = 14 Note: ⁿP_r = $$\frac{{{\text{n}}!}}{{\left( {{\text{n}} - {\text{r}}} \right)!}}$$

Solution:
The given word contains 8 letters of which E is taken 3 times.
∴ Required number of ways
$$\eqalign{ = \frac{{8!}}{{3!}} \cr = \frac{{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{6} \cr = 6720 \cr} $$

Solution:
The given words contains 8 letters out of which U is taken 2 times and all other letters are different.
∴ Required number of ways
$$\eqalign{ = \frac{{8!}}{{2!}} \cr = \frac{{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{2} \cr = 20160 \cr} $$