Use the formula for nCr to solve Exercises 49–56.

An election ba...

Question

Use the formula for nCr to solve Exercises 49–56.

An election ballot asks voters to select three city commissioners from a group of six candidates. In how many ways can this be done?

Accepted Answer

hey everyone in this problem we are asked to find how many possible combinations of three songs we can have if our album we're choosing these songs from have a total of 12 songs. So before we can solve this problem we need to first recall the possible combinations formula which is n C. R is equal to n factorial divided by the quantity of n minus R. R. Factorial times R factorial. Now in words this formula is the number of possible combinations if our items are taken from n items. So in this case our n items is the total number of songs we have in the album which is 12. And we are taking three songs from these songs so our our is going to be three. And now with this information identified we can go ahead and plug in these values into our formula. So we have 12 C. Three is equal to in the numerator instead of N factorial. We have 12 factorial. And in the denominator we have the quantity of N. Which is 12 minus R. Which is three factorial times R. Factorial which is just three factorial. And now we can begin simplifying. So in the numerator will keep the 12 factorial for now. And in the denominator we have 12 minus three which gives us nine factorial times three factorial. And since we have Have this nine factorial in the denominator we can expand the numerator until we also have nine factorial in the numerator. So doing this in the numerator we will have 12 times 11 times 10 times nine factorial. All divided by nine factorial times three factorial. And now with this we can see that our nine factorial clearly cancel out. So we can now write our simplified expression here and expand the denominator. So we have 12 times 11 times 10 divided by three times two times one. And now from here we can multiply out our denominator and simplify a bit more. So we have 12 times 11 times 10. All divided by three times two times one is just six. And now we have this six in the denominator. Well in the numerator we have 12. We can simplify this by dividing the 12 by by six giving us too Two in the numerator and one in the denominator. So now writing out this simplification here we are left with two times 11 times 10, multiplying these three terms together, we are left with the solution of 220. So now we know that we have a total possible number of combinations of 220 which is our answer here for choice. D Thanks for watching. I hope you found this video helpful and I'll see you again next time