Question

No. Of ways to select 5 people out of 12 people such that no two are consecutive

Answer 1

56

Answer 2

To select 5 people out of 12 such that no two are consecutive, we can use a standard combinatorial technique.

Let the 12 people be arranged in a row. Let $n=12$ be the total number of people and $k=5$ be the number of people to be selected.

Method 1: Using the transformation technique

Let the positions of the 5 selected people be $x_1, x_2, x_3, x_4, x_5$, such that $1 \le x_1 < x_2 < x_3 < x_4 < x_5 \le 12$. The condition that no two people are consecutive means that the difference between the positions of any two selected people must be at least 2. So, $x_{i+1} - x_i \ge 2$ for $i=1, 2, 3, 4$.

Let's define a new set of variables $y_i$:

$y_1 = x_1$

$y_2 = x_2 - 1$

$y_3 = x_3 - 2$

$y_4 = x_4 - 3$

$y_5 = x_5 - 4$

Now, let's check the conditions for $y_i$:

1. $y_1 \ge 1$ (since $x_1 \ge 1$)
2. $y_2 = x_2 - 1 \ge (x_1 + 2) - 1 = x_1 + 1 = y_1 + 1$. So, $y_2 > y_1$.
3. Similarly, $y_3 > y_2$, $y_4 > y_3$, and $y_5 > y_4$.

So, we have $1 \le y_1 < y_2 < y_3 < y_4 < y_5$.

What is the upper bound for $y_5$?

$y_5 = x_5 - 4 \le 12 - 4 = 8$.

Thus, we need to choose 5 distinct numbers ($y_1, y_2, y_3, y_4, y_5$) from the set $\{1, 2, 3, 4, 5, 6, 7, 8\}$. The number of ways to do this is given by the combination formula $\binom{m}{k}$, where $m$ is the total number of available choices (8 in this case) and $k$ is the number of items to choose (5 in this case).

Number of ways = $\binom{8}{5}$

Method 2: Using the gaps method

Imagine the $n-k$ unselected people are placed first. Number of unselected people = $12 - 5 = 7$. Let's represent the unselected people by 'U':

U U U U U U U

These 7 unselected people create $7+1=8$ possible positions (gaps) where the 5 selected people can be placed.

_ U _ U _ U _ U _ U _ U _ U _

To ensure no two selected people are consecutive, each selected person must be placed in a different gap. We need to choose 5 of these 8 available gaps for the 5 selected people.

The number of ways to choose 5 gaps out of 8 is $\binom{8}{5}$.

Calculation:

$\binom{8}{5} = \binom{8}{8-5} = \binom{8}{3}$

$\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56$

So, there are 56 ways to select 5 people out of 12 such that no two are consecutive.

General Formula:

For selecting $k$ items from $n$ items arranged in a row such that no two are consecutive, the number of ways is $\binom{n-k+1}{k}$.

In this problem, $n=12$ and $k=5$.

Number of ways = $\binom{12-5+1}{5} = \binom{8}{5} = 56$.

Explanation of the solution:

To select $k$ people from $n$ arranged in a row such that no two are consecutive, imagine placing the $n-k$ unselected people first. These $n-k$ people create $n-k+1$ possible slots where the $k$ selected people can be placed. Since no two selected people can be consecutive, each selected person must occupy a unique slot. Thus, the problem reduces to choosing $k$ slots out of $n-k+1$ available slots. For $n=12$ and $k=5$, this is $\binom{12-5+1}{5} = \binom{8}{5} = 56$.