In the specific case of 5C5, we are choosing 5 items out of a set of 5 items, which means we are choosing all of the items in the set. Since there is only one way to do this, the value of 5C5 is 1.

Solution. In both of our solving processes, we see that 5 C 2 = 10. In other words, there are 10 possible combinations of 2 objects chosen from 5 objects.

5P5 is the number of ways of picking 5 objects out of a group of 5 objects, where order matters. Whenever you select ALL of the objects and order matters, the formula for nPn is n! . Since 5! =5(4)(3)(2)(1)=120 , that answers the question at hand.

6C3 = the number of combinations of three one can choose from a pool of six unique items. So, it turns out, there are twenty ways to pick a set of three items from a pool of six unique items.

Both buttons are a way to erase or clear an entry. The CE (clear entry) button clears the most recent entry while the C (clear) button will clear all input to the calculator. So, if you are typing a long computation and make a mistake, press the CE button as it will delete just the last digit.

Combinations are a way to calculate the total outcomes of an event where order of the outcomes does not matter. To calculate combinations, we will use the formula nCr = n! / r! * (n - r)!, where n represents the total number of items, and r represents the number of items being chosen at a time.

As a general rule in the circle of fifths, when you move clockwise around the circle, you use the sharp key name for enharmonic pitches (for instance you'd use C# instead of Db), but when you move counter-clockwise, you use the flat key name (for instance you'd use Gb instead of F#).