Question 11/20: Set 3 - Try Out - Mathematical Reasoning (SNBT)

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Number 11

A conference is attended by $7$ countries: Armenia, Belgium, China, Denmark, Ecuador, Finland, and Ghana. The flags of each country will be flown on flagpoles arranged in a single row. How many ways can the $7$ flags be arranged so that Armenia and Belgium are at the ends?

$\frac{5!}{2}$

$5!$

$2(5!)$

$2(6!)$

$\frac{7!}{2}$

Explanation

Filling Slots Method

We have $7$ flags with a specific condition: Armenia and Belgium must be at the ends of the row. We can visualize the arrangement using slots (boxes).

Step 1: Determine End Positions

There are only $2$ possibilities for the ends: Armenia on the left and Belgium on the right, or vice versa. Thus, we write the number $2$ in the first box as the multiplier for the end variations.

Step 2: Fill Middle Positions

There are $5$ remaining flags (China, Denmark, Ecuador, Finland, Ghana) to fill the $5$ middle positions. We fill the remaining slots in descending order ( $5, 4, 3, 2, 1$ ) following the permutation principle.

Here is the visualization of the filled slots:

\begin{array}{|c|c|c|c|c|c|} \hline 2 & 5 & 4 & 3 & 2 & 1 \\ \hline \end{array}

The number $2$ represents the ways to arrange the ends.

The numbers $5, 4, 3, 2, 1$ represent the ways to arrange the $5$ middle flags ( $5!$ ).

Final Calculation

The total number of ways is the product of all numbers in the boxes.

\text{Total} = 2 \times (5 \times 4 \times 3 \times 2 \times 1)

\text{Total} = 2 \times 5!

\text{Total} = 2(5!)

Thus, there are $2(5!)$ ways to arrange the flags.

Command Palette

Number 11

Explanation