+0

# Fill In the Blanks

+2
54
4
+27

Fill in the blanks to make the equation true. (Each blank should contain an integer.)

May 7, 2023

#1
-2

Let the blanks be X, Y, and Z.

The answer is X=1, Y=−1, and Z=2.

The proof is as follows:

\begin{align*} C(n,k) &= \frac{n!}{k!(n-k)!} \ &= \frac{n(n-1)(n-2)\cdots(n-k+1)}{k!} \end{align*}

We can rewrite this as follows:

\begin{align*} C(n,k) &= \frac{n(n-1)(n-2)\cdots(n-k+1)}{k!} \ &= \frac{n(n-1)(n-2)\cdots(n-k+1)}{(k-1)!(n-k)} \ &= \frac{n(n-1)(n-2)\cdots(n-k+1)(n-k)}{(k-1)!(n-k)} \ &= \frac{n(n-1)(n-2)\cdots(n-k+1)(n-k)}{(k-1)!}\cdot\frac{(n-k)!}{(n-k)!} \ &= \frac{n(n-1)(n-2)\cdots(n-k+1)(n-k)}{(k-1)!} \ &= C(n-2,k-1)(n-k) \end{align*}

We can also rewrite this as follows:

\begin{align*} C(n,k) &= \frac{n!}{k!(n-k)!} \ &= \frac{(n-2)!(n-k+2)!}{k!(n-k)!} \ &= \frac{(n-2)!}{k!}\cdot\frac{(n-k+2)!}{(n-k)!} \ &= C(n-2,k)(n-k+2) \end{align*}

Therefore, we have the following equations:

\begin{align*} C(n,k) &= C(n-2,k-1)(n-k) \ C(n,k) &= C(n-2,k)(n-k+2) \end{align*}

Solving these equations for X, Y, and Z, we get X=1, Y=−1, and Z=2.

May 7, 2023
#2
+27
0

It says the answer is incorrect

May 7, 2023
#3
+27
0

\bump

May 7, 2023
#4
0

\bump

Guest May 8, 2023