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 = 56

Let x= 8 and y= 3

therefore 8C3 = 56

Let x= 8 and y= 5

Therefore 8C5 = 56

Let x = 56 and y = 55

Therefore 56C55 = 56

Hence 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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Is there a fast way of doing this....I feel like on an actual exam, having to go through the calculations of this will take too much time....Thanks!

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