There are k-2 members in a certain band, including Jim and Allan. Two

priyabaheti

Let's just Plug In The Answers (PITA). I like trying B and D.

B: k=9, so there are 7 members in the band.
How many ways are there to select two people who aren't Jim or Allan? Let's call the other members P, Q, R, S, and T. There are five of them and we need two.
5C2 = (5*4)/(2*1) = 10 Yay, that's what we wanted!
Answer choice B.

PQ
PR
PS
PT
QR
QS
QT
RS
RT
ST


ThatDudeKnowsPITA

Since Jim and Allan are not selected, the two members to attend the awards ceremony were selected from among (k - 2) - 2 = k - 4 members. This choice can be made in \(_{k - 4}C_2 = \frac{(k - 4)!}{2!\times (k - 6)!}\) ways. Recall that (k - 4)! = (k - 4)(k - 5)(k - 6)!. Let's substitute this expression for (k - 4)! in the numerator and simplify:

\(\Rightarrow\frac{(k - 4)!}{2!\times (k - 6)!}\)

\(\Rightarrow\frac{(k - 4)(k - 5)(k - 6)!}{2!\times (k - 6)!}\)

\(\Rightarrow\frac{(k - 4)(k - 5)}{2}\)

We are told that two members from among the members other than Jim and Allan can be chosen in 10 ways, so let's set the above expression equal to 10 and solve for k:

\(\Rightarrow\frac{(k - 4)(k - 5)}{2} = 10\)

\(\Rightarrow (k - 4)(k - 5) = 20\)

\(\Rightarrow k^2 - 5k - 4k + 20 = 20\)

\(\Rightarrow k^2 - 9k = 0\)

\(\Rightarrow k(k - 9) = 0\)

\(\Rightarrow k = 0\quad\text{or}\quad k - 9 = 0\)

\(\Rightarrow k = 0\quad\text{or}\quad k = 9\)

Notice that k cannot equal 0 because in that case, the band would contain k - 2 = -2 members, which is not possible. Thus, k must be equal to 9.

Answer: B

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