Let Arrangement be,
B G B G B G B
4 boys can be seated in 4! Ways
Girl can be seated in 3! Ways
Required number of ways,
= 4! × 3!
= 144

The question requires you to find number of outcomes in which at most 3 coins turn up as heads.
i.e., 0 coins turn heads or 1 coin turns head or 2 coins turn heads or 3 coins turn heads.
The number of outcomes in which 0 coins turn heads is,
⁶C₀ = 1 outcome.
The number of outcomes in which 1 coin turns head is,
⁶C₁ = 6 outcomes.
The number of outcomes in which 2 coins turn heads is,
⁶C₂ = 15 outcomes.
The number of outcomes in which 3 coins turn heads is,
⁶C₃ = 20 outcomes.
Therefore, total number of outcomes
= 1 + 6 + 15 + 20
= 42 outcomes.

In order to reach (5,6) covering shortest distance at same time man has to make 5 horizontal and 6 vertical steps.
The number of ways in which se steps can be taken is given by,
$$\eqalign{ & = \frac{{11!}}{{5! \times 6!}} \cr & = 462 \cr} $$

Two horses A and B, in a race of 6 horses . . . A has to finish before B.
If A finishes 1 . . . . . B could be in any of or 5 positions in 5 ways and or horses finish in 4! Ways, so total ways = 5 × 4!
If A finishes 2 . . . . . B could be in any of last 4 positions in 4 ways. But or positions could be filled in 4! ways, so total ways = 4 × 4!
If A finishes 3^rd . . . . . B could be in any of last 3 positions in 3 ways, but or positions could be filled in 4! ways, so total ways = 3 × 4!
If A finishes 4^th . . . . . B could be in any of last 2 positions in 2 ways, but or positions could be filled in 4! ways, so total ways = 2 × 4!
If A finishes 5^th . . . . B has to be 6th and top 4 positions could be filled in 4! ways.
A cannot finish 6^th, since he has to be ahead of B.
Therefore total number of ways:
= 5 × 4! + 4 × 4! + 3 × 4! + 2 × 4! + 4!
= 120 + 96 + 72 + 48 + 24
= 360

In word 'MATHEMATICS', we treat vowels AEAI as one letter.
Thus, we have MTHMTCS (AEAI).
Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and rest are different.
&re4; Number of ways of arranging se letters = $$\frac{{8!}}{{\left( {2!} \right)\left( {2!} \right)}}$$   = 10080
Now, AEAI has 4 letters in which A occurs 2 times and rest are different.
Number of ways of arranging se letters = $$\frac{{4!}}{{2!}}$$ = 12
&re4; Required number of words = (10080 x 12) = 120960

The situation is similar to placing 10 identical balls among 5 distinguishable boxes, where a box may have zero or more balls in it.

This case can be represented as arranging ten balls and (5 - 1) four walls in single row, which can be done in ¹⁴C₄ ways. (The balls placed between every successive pair of walls belong to one group) ¹⁴C₄ = 1001 ways.

Two students can be selected from 6 in ⁶C₂ =15 ways.
Therefore, teacher goes to zoo 15 times.

A triangle needs 3 points.
And polygon of 8 sides has 8 angular points.
Hence, number of triangle formed,
$$\eqalign{ & { = ^8}{{\text{C}}_3} \cr & = \frac{{8 \times 7 \times 6}}{{1 \times 2 \times 3}} \cr & = 56 \cr} $$

$$\eqalign{ & = \left( {{}^{75}{P_2} - {}^{75}{C_2}} \right) \cr & = \left\{ {\frac{{75!}}{{75! - 2!}} - \frac{{75 \times 74}}{2}} \right\} \cr & = \frac{{75!}}{{73!}} - \left( {75 \times 37} \right) \cr & = \frac{{75 \times 74 \times 73!}}{{73!}} - \left( {75 \times 37} \right) \cr & = \left( {75 \times 74 - 75 \times 37} \right) \cr & = 75 \times 37 \times \left( {2 - 1} \right) \cr & = \left( {75 \times 37} \right) \cr & = 2775 \cr} $$

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 = $$\frac{{{\text{n}}!}}{{\left( {{\text{n}} - {\text{r}}} \right)!}}$$