Consider the arrangement, Where we are interested in
*no of ways to arrange *the followings in a row
( EDIT- Typing mistake in values so please go after EDIT )
a) 7 '+''s & 4 '-''s such that '-''s dont appear side by side
Answer- ⁷ C 4
what i did-  was, To place '-''s in a way that 2 '-''s dont look together i placed 6'+''s spacing from starting to end & between them: _ +
_+ _ + _ + _ + _ + _, so basically there i could find 7 spaces on which i can place 4'-''s, which is the correct answer!
B)5 Girls and 5 Boys so that no 2 girls adjacent=  4![for boys]x5![for girls]
C) 10 boys & 10 girls such that no girl appear side by side-**
Answer: 11!x10!
here why cant we apply the same logic from a)? If we apply it goes like: there will be 11 spaces for 10 girls if we arrange 10 boys by spaces as done in a), so 10 girls can be fitted in 11 places should be 11C10!
Despite similarities of questions there is no sign of patterns in answers. Is there anything more needs to be equipped to see the 'Pattern'?
**

*

*EDIT
**
a) 6 '+''s & 4 '-''s such that '-''s dont appear side by side
Answer- ⁷ C 4
what i did-  was, To place '-''s in a way that 2 '-''s dont look together i placed 6'+''s spacing from starting to end & between them: _ +
_+ _ + _ + _ + _ + _, so basically there i could find 7 spaces on which i can place 4'-''s, which is the correct answer!
BUT
b) 10 boys & 10 girls such that no girl appear side by side-**
Answer: 11!x10!
here why cant we apply the same logic from a)? If we apply it goes like: there will be 11 spaces for 10 girls if we arrange 10 boys by spaces as done in a), so 10 girls can be fitted in 11 places should be 11C10!
Despite similarities of questions there is no sign of patterns in answers. Is there anything more needs to be equipped to see the 'Pattern'?
 A: Some of the answers appear wrong to start with, so it is no wonder that patterns can't be seen !
$5$ boys and $5$ girls: $-B-B-B-B-B-$
The girls can be positioned in the gaps in $\binom65 = 6$ ways, and permuted in $5!$ ways, so the answer should be $6\times5!5!\;\; or\;\; 6!5!$
The second answer is also wrong.
There are $8$ places available for the minuses, but since all minuses are identical (unlike the girls) the answer should be $\binom84$
$-plus-plus-plus-plus-plus-plus-plus-$
I hope you can now see some pattern, and decide for yourself whether the last answer is right or wrong ?
A: The fundamental difference in the two cases is the fact that all +'s are identical and all -'s are identical but each boy are is different. So is each girl. No one boy is exactly the same as any other boy. So an arrangement is required in the second case.
@true pointed out this fact already, but I think you misunderstood what he/she meant.
Hence, in the first case, you correctly deduced that considering 7 gaps between every plus _ + _ + _ + _ + _ + _ + _ we get the answer as $^7C_4$
Similarly, in the second case, considering 11 gaps between 10 boys, _ B _ B _ B _ B _ B _ B _ B _ B _ B _ B _, the first step towards the answer gives us $^{11}C_{10}$
However, its not over yet...
Since each boy is distinct and each girl is distinct they can arrange among themselves to give rise to a different arrangement, which is not the case with +'s or with -'s as every + is the same as every other +, but every boy is not the same as every other boy.
Hence, to the above obtained answer we mutiply one $10!$ for girls and another $10!$ for boys.
Thus, Required Answer $= ^{11}C_{10} \times 10! \times 10! = \frac{11!}{1! \times 10!} \times 10! \times 10! = 11! \times 10!$, which is the given answer.
