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nevsk [136]
3 years ago
9

A car dealer has 7 new sedans she wants to display but room for only 5 on the showroom floor.

Mathematics
1 answer:
aev [14]3 years ago
4 0

Answer:

B) 21

Step-by-step explanation:

There are 7 cars total and 5 slots to fill for the showroom selections. Put another way, there are 7-5 = 2 cars that don't make it to the showroom floor. Since this value (2) is smaller than the previous number of slots (5), it makes it easier to ask the question "how many ways are there to pick 2 cars that don't make it to the show room floor?" instead of "how many ways are there to pick 5 cars that make it to the showroom floor?". We'll get the same answer either way.

So we have 7 total to pick from for slot 1, and then 6 left over for slot 2. Making 7*6 = 42 permutations. Order does not matter, meaning that picking a red car and then a blue car is the same as picking a blue and then red. The count 42 has double counted, so we divide by 2 to sort this out

42/2 = 21 represents the number of combinations. We can see this if we use the nCr combination formula with n = 7 and r = 2

nCr = (n!)/(r!*(n-r)!)

7C2 = (7!)/(2!*(7-2)!)

7C2 = (7!)/(2!*5!)

7C2 = (7*6*5!)/(2*5!)

7C2 = (7*6)/2

7C2 = 42/2

7C2 = 21

a similar computation happens if you use n = 7 and r = 5.

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