Solution P & C - "no two being together"

1. In how many ways can 5 soldiers be selected from among 15 soldiers standing in a row such that no two of the selected soldiers are standing consecutively?

If, 's' stands for the selected soldier and 'n' stands for the soldier not selected, the way in which the five soldiers are selected such that no two soldiers are consecutive would look like one of the following...

s n n n s n s n n s n n s n n

n n n n s n s n s n s n s n n

s n n n s n n n s n n n s n s

or something like this.

Thus the number of ways the selection can be done is same as the number of ways in which 5 's' and 10 'n' can be arranged such that no two 's' are together.
So that no two 's' are together, we arrange the 10 'n' first. Now since the 10 'n' are identical, they can be arranged in only 1 way. With the 10 'n' in place, there are 11 positions for the 5 's' and since all the 's' are also identical, we just have to choose 5 out of these 11 positions and the answer is ¹¹C₅ ways.

2. There are 15 empty cages arranged in a row. In how many ways can a lion, a tiger, a leopard, a panther and a puma be put in five of the cages with 1 animal in each such that no two occupied cages are consecutive?

This question is also similar to the above except that all the empty cages are identical but the occupied cages are distinct. Thus the answer has to be ¹¹P₅. If this is not clear, the following are the ways the animals can be caged.....('e' stands for an empty cage and 'o' stands for an occupied cage)

o e e e o e o e e o e e o e e

e e e e o e o e o e o e o e e

o e e e o e e e o e e e o e o

Next, the 5 different animals could be arranged in the 5 'o' positions in 5! ways and thus the answer is ¹¹C₅ x 5! i.e. ¹¹P₅.

Now for a practice question on the above (would require a little more funda of arranging similar objects)...
In how many ways can the letters of the word MATHEMATICS be arranged such that no two vowels are together? (Ans: 1260 x 840, the answer is given as a numeric value and not in terms of P or C so that it does not give away clues of how to solve :-))

Next post of P&C would deal with more complex condition on the above, say, arrange 4 girls and 12 boys such that there are atleast 3 boys between any two girls...