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Question asks how many ways can you choose 3 items out of 5 = 5C3 = 10
Answer B
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==> You get 5C3=5C2=(5)(4)/2!=10.

The answer is b.
Answer: B
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If set A={a,b,c,d,e}, what is the number of subsets that must include only 3 elements?

\(= 5C2\)

\(= \frac{5!}{2! * 3!}\)

\(= 5 * 2\)

\(= 10\)

Hence, Answer is B
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10 is the wrong answer here; it must be option D.15.

3 element subset of A{a,b,c,d,e} = 5C3 + 5.

5 needs to be added up to consider {a,a,a} ; {b,b,b} ; {c,c,c} ; {d,d,d} ; {e,e,e} which are also the subsets of A.

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doesnt the combination formula count this by default?
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doesnt the combination formula count this by default?
Nope 5C3 will be: 10

abc ; abd ; abe ; acd ; ace ; ade ; bcd ; bce ; bde ; cde
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