We have to arrange 4 letters of 5 letters word IMAGE without repetition. So, total permutations are n^r = 5⁴ = 625.

We know, ⁿP_r = \(\frac{n!}{(n-r)!}\) and ⁿC_r = \(\frac{n!}{(n-r)! r!}\).
=> ⁿP_r / ⁿC_r = \(\frac{\frac{n!}{(n-r)!}}{\frac{n!}{(n-r)! r!}}\) = r!
=> ⁿP_r = ⁿC_r * r!

Total digits are 1-9 i.e.9. Even number means 2,4,6,8 at unit?s place.
So, 4 possible digits for unit place, after a digit is placed at unit place 8 digits are remaining.
8 ways to fill 1st position, 7 ways to fill 2nd position, 6 ways to fill th3 3rd position.
Hence, by fundamental principle of counting, total possible numbers are 8*7*6*4 = 1344.

There are 5 vowels in word EDUCATION. 5 vowels can be arranged in ⁵P₅ i.e. 5! ways. When all vowels are toger, 5 vowels toger form one letter and remaining 3 letters i.e. toger 4 letters can be arranged in ⁴P₄ i.e. 4! ways. Total possible arrangements are 5! * 4! = 120*24 = 2880.

We know, n! = n.(n-1).(n-2).(n-3)…..
6! = 6.5.4.3.2.1 = 720.

We know, ⁿC_r = \(\frac{n!}{(n-r)! r!}\).
ⁿC₀ = \(\frac{n!}{(n-0)! 0!} = \frac{n!}{n!(1)}\) = 1.

If re are 4 paths to travel from Delhi to Kanpur and person take one path n in coming back journey only 3 paths are remaining. So, by fundamental principle of counting, total number of ways a person can travel from Delhi to Kanpur and came back to Delhi via different path are 4 * 3 = 12.

Out of 5 males, 2 males can be selected in ⁵C₂ ways.
⁵C₂ = \(\frac{5!}{(5-2)! 2!} = \frac{5*4*3!}{(3)! 2!} = \frac{20}{2}\) = 10 ways.
Out of 3 females, 2 females can be selected in ³C₂ ways .
³C₂ = \(\frac{3!}{(3-2)! 2!} = \frac{3*2!}{(1)! 2!} = \frac{3}{1}\) = 3 ways.
So, by fundamental principle of counting we can select a group of 2 males and 2 females from family in 10*3 = 30 ways.

\(\frac{100}{10!} = \frac{1}{8!} + \frac{x}{9!}\).
We know, n! = n.(n-1). (n-2). (n-3) ????? = n(n-1)!
\(\frac{100}{10.9.8!} = \frac{1}{8!} + \frac{x}{9.8!}\)
=> \(\frac{100}{10.9} = \frac{1}{1} + \frac{x}{9}\)
=> \(\frac{10}{9} = 1 + \frac{x}{9}\)
=> \(\frac{x}{9} = \frac{10}{9} ? 1 = \frac{1}{9}\)
=> x=1.

There are 5 vowels in word EDUCATION. 5 vowels can be arranged in ⁵P₅ i.e. 5! ways. When all vowels are toger, 5 vowels toger form one letter and remaining 3 letters i.e. toger 4 letters can be arranged in ⁴P₄ i.e. 4! ways. Total possible arrangements are 5! * 4! = 120*24 = 2880.
So, when all vowels do not occur toger, total possible arrangements = 8! ? 2880 = 40320 ? 2880 = 37440.