1. Solve for the permutation of thof the word AMBASSADOR.Given:n=_ p=; q=__Solution:. Created by galoygeorge1. Math

1. Solve for the permutation of thof the word AMBASSADOR.Given:n=_ p=; q=_

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PROBLEM:

Solve for the permutation of the word AMBASSADOR.

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SOLUTION:

• To find the number of ways you arrange the letters of the word AMBASSADOR, we'll use the formula for distinguishable permutation.

• Since there are 10 letters in total and A has been used thrice while S has been used twice, our given will be n = 7, p = 3 and q = 2.

$\huge \boxed{ \large \begin{array}{l} \rm P=\frac{n!}{p!q!} = \frac{10!}{3!2!}\\ \rm P= \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times4\times \cancel{3 \times 2 \times 1}}{ \cancel{3 \times 2 \times 1} \times 2 \times 1 }\\ \rm P= \frac{604,800}{2} \\ \rm P= 302,400 \end{array}}$

ANSWER:

• Thus, there are 302, 400 distinguishable permutations of the letters in the word AMBASSADOR.

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Question :

1. Solve for the permutation of the word AMBASSADOR.

Solving :

${\tt{P = \frac{n!}{p!q! } = \frac{10!}{3!2!}}} \\{\tt{P = \frac{10!}{3!2!}}} \\{\tt{P = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3}{3!2!}}} \\ {\tt{P = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times \cancel{3}}{\cancel{3!} 2!}}} \\{\tt{P = \frac{604.800}{2}}} \\{\tt{P = \frac{\cancel{604.800}}{\cancel{2}}}} \\ {\tt{ answer \: is :}} \\{\tt{P =\boxed{\tt{302.400}}}}$

${\tt{there \: are \: 302, 400}} \\ {\tt{ distinguishable \: permutation}}$

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