Answer:
4
Step-by-step explanation:
6^(7-3) = 6^4
that is the answer
Answer:
Step-by-step explanation:
I should know, I've done this.
The are 40320 ways in which the 5 indistinguishable rooks be can be placed on an 8-by-8 chess- board so that no rook can attack another and neither the first row nor the first column is empty
<h3 /><h3>What involves the
rook polynomial? </h3>
The rook polynomial as a generalization of the rooks problem
Indeed, its result is that 8 non-attacking rooks can be arranged on an 8 × 8 chessboard in r8.
Hence, 8! = 40320 ways.
Therefore, there are 40320 ways in which the 5 indistinguishable rooks be can be placed on an 8-by-8 chess- board so that no rook can attack another and neither the first row nor the first column is empty.
Read more about rook polynomial
brainly.com/question/2833285
#SPJ4
Answer:
6xrx7
2
Regroup terms.
6x{x}^{7}r6xx7r
3
Use Product Rule: {x}^{a}{x}^{b}={x}^{a+b}xaxb=xa+b.
6{x}^{1+7}r6x1+7r
4
Simplify 1+71+7 to 88.
6{x}^{8}r6x8r
Answer:
9,240
Step-by-step explanation:
For the first number, there are 22 numbers available to choose from. For the second number, there are 21 numbers available, since the first number cannot be the same as the second number. And for the last number, there are 20 remaining numbers to choose from. So, there are 22*21*20=9240 possible sequences.
