Exploring Combinations and Permutations: Marlon's Art Museum Adventure

Marlon is working at the local art museum and has a challenging task. He has ten pieces of art, but only four frames to display. The question is, how many different ways can Marlon select and arrange four frames from these ten pieces of art? This problem delves into the realms of permutations and combinations, fundamental concepts in mathematics with wide-ranging applications in real-world scenarios.

Understanding the Problem

Marlon has ten pieces of art, but he can only hang four of them. To solve this problem, we need to differentiate between two scenarios: selecting four frames from ten and arranging them, or simply selecting four frames without specifically arranging them. The number of ways to select and arrange the four frames is a permutation problem, while the number of ways to select the four frames without arranging them is a combination problem.

Permutations: Arranging Selected Art Pieces

Let's start with permutations, which involve arranging items in a specific order. Marlon can choose any of the ten pieces of art for the first frame, any of the remaining nine pieces for the second frame, any of the remaining eight pieces for the third frame, and any of the remaining seven pieces for the fourth frame. We can calculate this as follows:

10 × 9 × 8 × 7 5,040 ways to choose and arrange four frames from ten pieces of art.

Combinations: Selecting Art Pieces Without Arranging

Now, let's consider combinations, where the order of selection does not matter. Here, Marlon needs to select four pieces of art from the ten, and the selection process does not consider the order. The formula for combinations is given by:

[ C(n, k) frac{n!}{k!(n-k)!} ]

Where ( n ) is the total number of items, and ( k ) is the number of items to choose. For Marlon's case, ( n 10 ) and ( k 4 ).

[ C(10, 4) frac{10!}{4!(10-4)!} frac{10!}{4!6!} ]

Calculating the factorial values:

[ 10! 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 3,628,800 ]

[ 4! 4 × 3 × 2 × 1 24 ]

[ 6! 6 × 5 × 4 × 3 × 2 × 1 720 ]

Substituting these values into the combination formula:

[ C(10, 4) frac{3,628,800}{24 × 720} frac{3,628,800}{17,280} 210 ]

So, Marlon can select four frames from ten pieces of art in 210 different ways.

Additional Question: Arranging Remaining Art Pieces

Suppose Marlon has already selected and arranged four frames. Now, he has six remaining pieces of art. How many ways can he select and arrange four frames from these six pieces of art?

The same principle of permutations applies here. The number of ways to choose and arrange four frames from six pieces of art is calculated as:

6 × 5 × 4 × 3 360 ways.

Alternatively, using the combination formula:

[ C(6, 4) frac{6!}{4!(6-4)!} frac{6!}{4!2!} frac{720}{24 × 2} frac{720}{48} 15 ]

Marlon can select and arrange four frames from the remaining six pieces of art in 15 ways.

Conclusion

Marlon's art museum task is a perfect example of how mathematical concepts like permutations and combinations can be applied in real-world scenarios. Understanding these concepts helps in accurately solving such problems.

Related Keywords: permutations, combinations, art museum