The number of times the teacher goes to the zoo $$ = {\,^n}{C_3}.$$
The number of times a particular child goes to the zoo $$ = {\,^{n - 1}}{C_2}.$$
From the question, $$^n{C_3} - {\,^{n - 1}}{C_2} = 84$$
or, $$\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right) = 6 \times 84 = 9 \times 8 \times 7$$
$$ \Rightarrow \,\,n - 1 = 9.$$

The first place from the left can be filled in 9 ways (any one except 0).
The other eight places can be filled by the remaining 9 digits in $$^9{P_8}$$ ways.
∴ the number of 9-digit numbers $$ = 9{ \times ^9}\,{P_8}.$$

$$\eqalign{ & L = {\,^{10}}{C_2} - {\,^4}{C_2} + 1 = 45 - 6 + 1 = 40 \cr & T = {\,^{10}}{C_3} - {\,^4}{C_3} = 120 - 4 = 3\,L - 4 \cr} $$

Total number of ways of arranging the letters of the word $$BANANA$$   $$\frac{{6!}}{{2!3!}} = 60.$$   Number of words in
which $$2\,N’s$$  come together is $$\frac{{5!}}{{3!}} = 20.$$
Hence the required number $$= 60 - 20 = 40$$

Possible cases for $$X$$ are
(1) 3 ladies, 0 man
(2) 2 ladies, 1 man
(3) 1 lady, 2 men
(4) 0 ladies, 3 men
Possible cases for $$Y$$ are
(1) 0 ladies, 3 men
(2) 1 lady, 2 men
(3) 2 ladies, 1 man
(4) 3 ladies, 0 man
No. of ways $$^4{C_3}.{\mkern 1mu} {{\mkern 1mu} ^4}{C_3} + {\left( {^4{C_2}.{\mkern 1mu} {{\mkern 1mu} ^3}{C_1}} \right)^2} + {\left( {^4{C_1}.{\mkern 1mu} {{\mkern 1mu} ^3}{C_2}} \right)^2} + {\left( {^3{C_3}} \right)^2}$$
$$= 16 + 324 + 144 + 1 = 485$$

$$\eqalign{ & 62 = {\,^n}{C_1} + {\,^n}{C_2} + {\,^n}{C_3} + ..... + {\,^n}{C_{n - 1}} = {2^n} - {\,^0}{C_0} - {\,^n}{C_n} = {2^n} - 2 \cr & \therefore \,\,{2^n} = 64 = {2^6}. \cr} $$

4 novels, out of 6 novels and 1 dictionary out of 3 can be selected in $$^6{C_4} \times {\,^3}{C_1}$$   ways
Then 4 novels with one dictionary in the middle can be arranged in 4! ways.
∴ Total ways of arrangement = $$^6{C_4} \times {\,^3}{C_1} \times 4! = 1080$$

The matrix will be of the order $$4 \times 1$$  or $$1 \times 4$$  or $$2 \times 2.$$
For each order, the number of different matrices
= the number of ways to fill four places by 0, 1, 2, 3
$$ = 4 \times 4 \times 4 \times 4.$$

\[ \times \times \left| {\begin{array}{*{20}{c}} {1,3,5} \\ \times \end{array}} \right.\left| {\begin{array}{*{20}{c}} {2,6} \\ \times \end{array}} \right.\]
The units place can be filled in 2 ways.
The tens place can be filled in 3 ways.
The remaining two places can be filled by any two of the remaining three digits. So, the required number of numbers $$ = 2 \times 3 \times {\,^3}{P_2}.$$

First let us arrange $$M, I, I, I, I, P, P$$
Which can be done in $$\frac{{7!}}{{4!2!}}$$   ways
$$\sqrt {{M}} \sqrt {{I}} \sqrt {{I}} \sqrt {{I}} \sqrt {{I}} \sqrt {{P}} \sqrt {{P}} $$
Now $$4S$$ can be kept at any of the ticked places in $$^8{C_4}$$  ways so that no two $$S$$ are adjacent.
Total required ways
$$\frac{{7!}}{{4!2!}}{\,^8}{C_4} = \frac{{7!}}{{4!2!}}{\,^8}{C_4} = 7 \times {\,^6}{C_4} \times {\,^8}{C_4}$$