Solution:
This problem is a permutation because the order matters here. This means that choosing A as King, B as Knight, C as Bishop and D as Rook results in a different arrangement from B as King, A as Knight, D as Bishop and C as Rook. We would count them both because in the first case A is King, but in the second case A is Knight.
Therefore, the possible number of ways are given below:
![19P4=\frac{19!}{(19-4)!} =\frac{19 \times 18 \times 17 \times 16 \times 15!}{15!} =19 \times 18 \times 17 \times 16 =93024](https://tex.z-dn.net/?f=19P4%3D%5Cfrac%7B19%21%7D%7B%2819-4%29%21%7D%20%3D%5Cfrac%7B19%20%5Ctimes%2018%20%5Ctimes%2017%20%5Ctimes%2016%20%5Ctimes%2015%21%7D%7B15%21%7D%20%3D19%20%5Ctimes%2018%20%5Ctimes%2017%20%5Ctimes%2016%20%3D93024)
Hence there will be 93024 ways 19 members of a chess club fill the offices of King, Knight, Bishop, and Rook.
Answer:
the answer is d (x+4)^2/9 - (y-1)^2/25
Step-by-step explanation:
its correct on edge
Answer:
The correct option is option (B)
Yes, he will have enough poster board because 10 = 2.5 feet. Then, 2.5 x 2 = 5 square feet .
Step-by-step explanation:
1 foot = 12 inches.
Fabian needs 45 square feet of poster board for an art.
Actually, the area of the poster is 4.5 square feet.
[Since the unit of 4.5 is square feet, so it is unit of area.]
He buys a sheet of poster board with length 30 inches and a width 2 feet.
The units of length and width are not same.
So, first we need to convert the width in inches.
The length of the sheet is = 30 inches
=(30÷12) inches
=2.5 feet.
Total area of the sheet = length×width
=(2.5×2) square feet
= 5 square feet
So, the area of sheet is more than the area of poster.
Yes, he will have enough poster board because 10 = 2.5 feet. Then, 2.5 x 2 = 5 square feet .