Apaint box contains 12 bottles of different colors. if we choose equal quantities of 3 different colors at random, how many color combinations are possible? a. 479,001,600 b. 1,320 c. 220 d. 36

C

Step-by-step explanation:

You are choosing 3 from a total of 12. Order does not matter. So you are working with combinations. The answer symbolically is

12C3

12C3 = 12!/(9!3!)

12C3 = 12 * 11 * 10 * 9!/(9! 3!)

12C3 = 12 * 11 * 10/6

12C3 = 2 * 11 * 10

12C3 = 220

We have 12 choices for the first selection, 11 for the second, and 10 for the third. There are 12*11*10 = 1320 permutations. If your teacher was asking about permutations, then you would be done at this point. However, your teacher is asking about combinations. With combinations, order does not matter.

For any group of 3 items, there are 3! = 3*2*1 = 6 ways to arrange this group. This means that we must divide 1320 over 6 to correct for the fact that we overcounted by a factor of 6

In this case,

number of combinations = (number of permutations)/6

number of combinations = 1320/6

number of combinations = 220

More generally, I'm using the connection that

nCr = (nPr)/(r!)

step-by-step explanation:

a.   since the profit made in 2012 is equal to the profit made in 2014, the profit is           symmetric about the line y = 120,000.

step-by-step explanation:

given that megan owns a clothing boutique.

the profits of 2010, 11, are recorded and shown as follows:

years since     2010 0     1     2     3       4       5

profits made in 1000's   39       84     111     120   111     84

we find that profit gradually increases form 2010 to 2013 upto 120,   then starts decreasing.

the profits for 2012 and 2014 are the same while that of 2011 and 2015 are also the same.

this shows that there is symmetry about the line profit = 120000

hence answer is:

a.   since the profit made in 2012 is equal to the profit made in 2014, the profit is           symmetric about the line y = 120,000.

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the equivalent expression are always the hardest ones but you look at the gcf also

step-by-step explanation: