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3 items x, y and z will have 6 different permutations however only one combination. The given formular is generally used to determine the number of combinations in a described situation.
nCr = (n!)/((r!)(n - r)!)
Solution
i. 8C7 = (8!)/((7!)(8 - 7)!)
= 8!/7! 1!
= (8 * 7!)/1 * 7!
= 8
ii. 6C4 = (6!)/((4!)(6 - 4)!)
= 6!/4! 2!
= (6 * 5 * 4!)/(4! * 2 *1)
= 15
iii. 8C3 = (8!)/((3!)(8 - 3)!)
= (8 *7 *6 *5!)/(3 * 2 * 1 * 5!)
= 56
solve x+y= 7 and x-y =21
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