Let us first seat  the 5 girls (or 5 boys). This can be done in 5! ways.
Now, for each such arrangement, the 5 boys can be seated only at the cross marked the place as shown below:
(I)  \(\times G_{ 1\times }G_{ 2 }\times { G }_{ 3 }\times { G }_{ 4 }\times { G }_{ 5 }\times \)
(II) \(G_{ 1\times }G_{ 2 }\times { G }_{ 3 }\times { G }_{ 4 }\times { G }_{ 5 }\times \)
In case I, 5 boys can be seated in P = 5! ways
Thus, number of ways of seating = 5! \(\times \) 5! = \({ (5! })^{ 2 }\)
Similarly, in case II, number of ways of seating = \({ (5! })^{ 2 }\) 
Hence, required number of ways = \({ (5! })^{ 2 }\) + \({ (5! })^{ 2 }\) \({ =2.(5! })^{ 2 }=28800\)
For solving this type of problem, we used the following steps
Step I Firstly, decide that from how many digits the required number will be formed.
Step II Fill up the places on which restrictions are present and let the number of ways of filling up these places be k.
Step III Find the number of ways of filling the remaining places with remaining digits by using the formula ⁿP _r  .
Step IV Required number of numbers is k. P . ⁿP _r .

We have 5 boys and 5 girls,
Since no two girls sit together, therefore the possible choices for girls are the places marked as '\(\times\)'.
B1  \(\times\)B2   \(\times\)B3 \(\times\)B4 \(\times\)B5 \(\times\)
Clearly, the girls can be arranged in ⁶P5 ways and the boys can be arranged in 5! ways.
Hence, by fundamental principle multiplication, required number of ways = ⁶P_{5 \(\times \)5!}
 \(=\frac { 6! }{ (6-5)! } = 5!=6!\times 5!=86400\)
For solving this type of problem, we used the following steps
Step I Firstly, decide that from how many digits the required number will be formed.
Step II Fill up the places on which restrictions are present and let the number of ways of filling up these places be k.
Step III Find the number of ways of filling the remaining places with remaining digits by using the formula  ⁿP_r
Step IV Required number of numbers is k. ⁿ P _r .

Ans . 41472

(i) Let us take 2 particular children together as one. Now, the remaining 4 (particular children's and other three children's) can be arranged in 4! = 24 ways. Again two particular children taken together can be arranged in 2! = 2 ways.
Hence, there are 24 x 2 = 48 ways of arrangement.
(ii) Clearly, required number of arrangements = Number of permutation of 5 children taken all at a time - Number of permutation of children in which two particular children are together
= 5! - 4! x 2 = 5 x 4! - 4! x 2 
= 4! (5-2) = 24 x 3 = 72
Hence, required number of arrangements = 72

There will be as many ways as there are ways of filling 2 vacant places  in succession by the five given digits. Here, in this case, we start filling in unit’s place, because the options for this place are 2 and 4 only and this can be done in 2 ways; following which the ten’s place can be filled by any of the 5 digits in 5 different ways as the digits can be repeated. Therefore, by the multiplication principle, the required number of two digits even numbers is 2 × 5, i.e., 10.