Selections and Combinations.

stars and bars technique

Let

• bars | be the 5 numbers to be chosen, total 5 bars.
• stars * be the remaining numbers from the first 15 natural numbers, total 10 stars.

example:

$*|*|**|***|*|**$

Count from left to right, the selected numbers are $2, 4, 7, 11, 13$

Construct the solution:

As no two of them are consecutive, we first put a star in-between the bars

$|*|*|*|*|$

Then there are remains 6 stars to be placed in 6 sections (urns).

<urn 1> $|*$ <urn 2> $|*|*|*|$ <urn 6>

one of those solutions as example :

$\color{#3D99F6}***$ $|*$ $\color{#3D99F6}***$ $|*|*|*|$

the chosen numbers are $4,9,11,13,15$

The number of required selections

= the number of ways to arrange 6 stars into 6 urns

= ${6+6-1} \choose {6-1 }$

$= \cfrac{11 \times 10 \times 9 \times 8 \times 7}{1 \times 2 \times 3\times 4 \times 5} = 33 \times 14$

$\therefore k = \boxed{14}$

n(choosing 5 non-consecutive numbers from 15 consecutive ones)
= (15 – 5 + 1)C5
= 11C5
= 11! / 5!6!
= 33k

k = 14

There are four spaces between the five chosen numbers. We cannot choose from these four spaces. Therefore, we have (15-4) choose 5 possibilities, which is equal to 33*14.

IF we define $b_j = a_j - j$ for $1 \le j \le 5$ , then $a_1,a_2, ..., a_5$ is an increasing sequence of nonconsecutive integers between $1$ and $15$ if and only if $b_1, b_2, ...., b_5$ is an increasing sequence of integers between $0$ and $10$ . Thus we want the number of ways of choosing $5$ distinct numbers from $11$ , namely $\binom{11}{5} = 462 = 33 \times \boxed{14}$ .