Answer:
26
Step-by-step explanation:
Data provided in the question:
set {1, 2, 3, 5, 11}
Now,
Total number of different choices of a number available = 5
Therefore,
Number of ways to choose 2 distinct numbers= ⁵C₂
Number of ways to choose 3 distinct numbers= ⁵C₃
Number of ways to choose 4 distinct numbers= ⁵C₄
Number of ways to choose 5 distinct numbers= ⁵C₅
therefore,
Total number we can get
= ⁵C₂ + ⁵C₃ + ⁵C₄ + ⁵C₅
= [tex]\frac{5!}{2!(5-2)!}+\frac{5!}{3!(5-3)!}+\frac{5!}{4!(5-4)!}+\frac{5!}{5!(5-5)!}[/tex]
= [tex]\frac{5\times4\times3!}{2!3!}+\frac{5\times4\times3!}{3!\times2!}+\frac{5\times4!}{4!\times1!}+\frac{5!}{5!\times0!}[/tex]
= 10 + 10 + 5 + 1
= 26
Answer:
15
Step-by-step explanation:
The answer is actually 15 because 2*3*5*11, meaning 4+6+4+1, which equals 15. (This is an AOPS question, so I know this is right)
