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charle [14.2K]
3 years ago
8

How many different ways are there to arrange the letters A, B, A, and B? 4 6 8 16

Mathematics
1 answer:
andriy [413]3 years ago
6 0

There are 6 different possible arrangements of letters A, B, A, B.

<u>Solution: </u>

Need to determine different ways to range letters A, B, A and B.

Using the theorem which says that the number of permutation of n alphabets, where p_1 number of alphabets of one kind and p_2 is number of alphabets of second kind is given by following formula.

Number of possible arrangements =\frac{n !}{p_{1} ! \times p_{2} !} \rightarrow(1)

In our case total number of alphabets = n = 4

Number of letter A = p_1 = 2

Number of letter B = p_2 = 2

Using (1), we get

Number of possible arrangements of A, B, A, B =\frac{4 !}{2 ! \times 2 !}=\frac{4 \times 3 \times 2 \times 1}{2 \times 1 \times 2 \times 1}=3 \times 2=6

Hence there are 6 different possible arrangements of letters A, B, A, B.

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The solution to the division of the given surd is: \mathbf{P =\dfrac{(6\sqrt{x}-x^2\sqrt{x}-3x\sqrt{x}+2x+2)(x-1) }{8x}      }

<h3>Division of Surds.</h3>

The division of surds follows a systemic approach whereby we divide the whole numbers separately and the root(s) are being divided by each other.

Given that:

\mathbf{P=(\frac{\sqrt{x} +2}{x-1}+\frac{\sqrt{x}\\ -2}{x-2\sqrt{x} +1} ) : \frac{4x}{(x-1)^{2} }}

i.e.

\mathbf{=\dfrac{(\frac{\sqrt{x} +2}{x-1}+\frac{\sqrt{x}\\ -2}{x-2\sqrt{x} +1} )}{ \frac{4x}{(x-1)^{2} }} }

Using the fraction rule:

\mathbf{\dfrac{a}{\dfrac{b}{c}}= \dfrac{a\times c}{b}}

\mathbf{\implies \dfrac{(\frac{\sqrt{x} +2}{x-1}+\frac{\sqrt{x}\\ -2}{x-2\sqrt{x} +1} )(x-1)^{2}}{4x}} }

By simplification, we have:

\mathbf{  =\dfrac{\dfrac{(6\sqrt{x}-x^2\sqrt{x}-3x\sqrt{x}+2x+2)(x-1) }{2}  }{4x}      }

\mathbf{P =\dfrac{(6\sqrt{x}-x^2\sqrt{x}-3x\sqrt{x}+2x+2)(x-1) }{8x}      }

Learn more about evaluating the division of surds here:

https://brainly.in/question/27942899

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