Last answered:

26 May 2020

Posted on:

12 May 2020

0

Girlfriend Vs Boyfriend Combinatorics and Variations

YOU can help end a dispute between a girlfriend and boyfriend in Quarantine!
We both tried calculating our own ways and we both came to no conclusion and only confusion.
I tried putting my course knowledge to the test to impress my girlfriend and I have gotten lost in a web of formulas.
Here is our word math problem!

We want to know how many COMBINATIONS you can achieve with JUST your 5 digit fingers WITH REPETITION and ORDER DOES NOT MATTER.
Example Combination: Pointer Finger and Thumb; Pointer Finger and Middle Finger; Pinky, Thumb, and Middle Finger; Thumb, Pointer, Middle, Ring....etc
The combos can be from 2 to all 5 fingers. 

Combinations with Repetition:
= (n + p - 1)! / p! (n – 1)!

From the course I looked back at my notes and was trying to plug the above.
I assume "n" would be our digits, so I inputted 5 there. The "p", I'm finding to have trouble what to input there...

Please ease tension in our home! But also! Help me crack this code.

1 answers ( 0 marked as helpful)
Instructor
Posted on:

26 May 2020

0
Hey Bryan,    Sorry for the late response, but I've had some medical issues and I wasn't able to work for several weeks. :/ Now that I'm feeling better, I'm here to resolve this fierce family debate!   We want to know how many COMBINATIONS you can achieve with JUST your 5 digit fingers WITH REPETITION and ORDER DOES NOT MATTER.   Alright! So, you're misunderstanding the concept of repetition here. In terms of variations, combinations and the like, repetition implies that the same object (finger in this case) can be used repeatedly in the same "group". For example, a combination being "Thumb - Thumb", which doesn't really work. Hence, we'll be relying on combinations without repetition.    Now that this is clear, you need to understand how combinations formula works. In it, we're picking a fixed number of elements out of a sample space with fixed size. In your case, we're not picking a fixed number of elements. We can pick 1, 2, 3, 4 or even 5 fingers, right?   So what do we do in this scenario? Well, we find the combinations of using 2, 3, 4 and 5 fingers and we add them all up. C(1;5) + C(2;5) + C(3;5) + C(4;5) + C(5;5) = 5!/(1!4!) + 5!/(2!3!) + 5!/(3!2!) + 5!/(4!1!) + 5!/(5!0!) = 5 + 10 + 10 + 5 + 1 = 31.    Of course, depending on whether choosing a single finger and/or all fingers is part of what you're looking for, you might exclude C(1;5) and/or C(5;5) from the calculation.    I hope this settles your family feud. :D    Best,  365 Vik

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