Four cats and five mice enter a race. In how many ways can they finish with a mouse placing first, second, and third

7200 ways

Step-by-step explanation:

Four cats and five mice enter a race. In how many ways can they finish with a mouse placing first, second, and third

Since order is important, we use combination rule

The mouse has to come First, Second and Third hence,

= C(n, r) = nCr = 5C3

3 mice coming 1st, Second and Third =

3! × 5C3

= 3! × 5!/3!(5 - 3)!

= 3! × 5!/3! × 2!

= 3! × 10

Then we have left 4 cats and 2 mice

= (4 + 2)!

= 6 !

Hence,

= 3! × 10 × 6!

= 7200 ways

3-4

step-by-step explanation:

the answer is 3-4 because both 45 and 60 has a multiple of 15.

45/3=15

60/4=15

therefore the odds against selecting a red is 4-3

d, x=8

step-by-step explanation:

simplifying

4x + -5 = 2x + 11

reorder the terms:

-5 + 4x = 2x + 11

reorder the terms:

-5 + 4x = 11 + 2x

solving

-5 + 4x = 11 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-5 + 4x + -2x = 11 + 2x + -2x

combine like terms: 4x + -2x = 2x

-5 + 2x = 11 + 2x + -2x

combine like terms: 2x + -2x = 0

-5 + 2x = 11 + 0

-5 + 2x = 11

add '5' to each side of the equation.

-5 + 5 + 2x = 11 + 5

combine like terms: -5 + 5 = 0

0 + 2x = 11 + 5

2x = 11 + 5

combine like terms: 11 + 5 = 16

2x = 16

divide each side by '2'.