Answer:
4989600 ways
Step-by-step explanation:
From the question,
The word MATHEMATICS can be arranged in n!/(r₁!r₂!r₃!)
⇒ n!/(r₁!r₂!r₃!) ways
Where n = total number of letters, r₁ = number of times M appears r₂ = number of times A appears, r₃ = number of times T appears.
Given: n = 11, r₁ = 2, r₂ = 2, r₃ = 2
Substitute these value into the expression above
11!/(2!2!2!) = (39916800/8) ways
4989600 ways
Hence the number of ways MATHEMATICS can be arranged without duplicate is 4989600 ways
Answer: 3X+4 is the correct answer
Step-by-step explanation:
There an infinite number of solutions that can be generated by giving different values
Let Bea = x
Therefore Andrew = x + 2
Chris = x - 3
Andrew * Chris = 66
(x + 2)(x - 3) = 66 Remove the brackets.
x^2 + 2x - 3x - 6 = 66
x^2 - x - 6 = 66 Subtract 66 from both sides.
x^2 - x - 72 = 0
(x - 9)(x + 8) = 0
Find the possible values of Bea's age
x - 9 = 0
x = 9 That value works. She can be plus 9 years old.
x + 8 = 0
x = - 8. There's no way she can be - 8 years old
So Bea is 9.
9 <<<<<< Answer
