On a TV game show, a contestant is shown 8 products from a grocery store and is asked to choose the three least-expensive items in the set, and then correctly a

Question

On a TV game show, a contestant is shown 8 products from a grocery store and is asked to choose the three least-expensive items in the set, and then correctly arrange these three items in order of price. In how many ways can the contestant choose the three items?

Answer 1

This questions is asking about permutations, which is the number of different ways to arrange a set of objects/things/etc. Permutations use the factorial function (!). In general, n!=n(n-1)(n-2)... If we have 4!, then we have 4*(4-1)*(4-2)*(4-3)= 4*3*2*1=24

For permutations less than all, permutations with more than one event/step, the following formula is used: nPk, with n being a superscript and k being a subscript. n represents the first event and k represents the second. This can be represented as:

nPk=n(n-1)(n-2)...to k factors.

Written in factorial form: n!/(n-k)!

For this problem, the first event is picking 3 objects from the total of 8, and then from the 3 you pick you need to arrange them in order of price. The first event involved 8 items (n=8) and the second event involved 3 items (k=3)

8P3=8*(8-1)*(8-2)=8*7*6=336

in factorial form: 8!/(8-3)! = 8!/5!

in the expanded form: (8*7*6*5*4*3*2*1)/(5*4*3*2*1).

5, 4, 3, 2, and 1 are the same in the numerator and denominator so they can cancel and you are left with: 8*7*6