of the 10 contestants in a televised quiz competition, 2 were eliminated during this weeks show, and 2 more…

of the 10 contestants in a televised quiz competition, 2 were eliminated during this weeks show, and 2 more will be eliminated during next weeks show. how many groups of 6 remaining contestants are possible a week from now? assume that the show airs on the same day every week.\n28\n210\n20,160\n151,200

of the 10 contestants in a televised quiz competition, 2 were eliminated during this weeks show, and 2 more will be eliminated during next weeks show. how many groups of 6 remaining contestants are possible a week from now? assume that the show airs on the same day every week.\n28\n210\n20,160\n151,200

Answer

Explanation:

Step1: Calculate remaining contestants

Initially 10 contestants. 2 are eliminated this week and 2 next week. So remaining contestants are $10 - 2 - 2=6$.

Step2: Use combination formula

We want to find the number of groups of 6 from 6 contestants. The combination formula is $C(n,r)=\frac{n!}{r!(n - r)!}$, where $n = 6$ and $r = 6$. So $C(6,6)=\frac{6!}{6!(6 - 6)!}=\frac{6!}{6!0!}$. Since $0!=1$, we have $C(6,6) = 1$. But if we assume the question means choosing 6 from the initial 10 before any eliminations (a more likely interpretation as the calculation above is trivial), then $n=10-(2 + 2)=6$ and we want to choose groups of 6. Using the combination formula $C(n,r)=\frac{n!}{r!(n - r)!}$ with $n = 6$ and $r=6$, we get $C(6,6) = 1$. If we consider choosing 6 from the initial 10 contestants before eliminations, $n = 10$ and $r=6$. Then $C(10,6)=C(10,4)$ (because $C(n,r)=C(n,n - r)$). And $C(10,4)=\frac{10!}{4!(10 - 4)!}=\frac{10\times9\times8\times7}{4\times3\times2\times1}=210$.

Answer:

B. 210