Bunuel wrote:
A group of x applicants is to be chosen for a spot on a television show. The number of applicants chosen, y, can range from 1 to x. If there are a total of 56 different possible selections for the spot on the television show, which of the following are possible values of y? (Choose all that apply.)
Now we have xCy = 56
where y= 1 to x
Let us consider the options which are mentioned whose combination equals to 56
56 = 7 * 8 or 14 * 4 or 28 * 2 or 56 * 1
Let x= 56 and y = 1
therefore 56C1 = 56Let x= 8 and y= 3
therefore 8C3 = 56Let x= 8 and y= 5
Therefore 8C5 = 56Let x = 56 and y = 55
Therefore 56C55 = 56Hence when y= 1 , 3, 5 and 55 then the combination equals to 56
i. e
option A, option B, option C and option G
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