Question

(1) STATISTICS
(2) BOOKKEEPER
(3) APPEARING
It
11. How many total arrangements can be made using all letters of the following words​

Answer 1

$\large\underline\blue{\bold{Given \:  Question :-  }}$

How many total arrangements can be made using all letters of the following

$\begin{gathered}\begin{gathered}\bf words= \begin{cases} &\sf{1. STATISTICS} \\ &\sf{2. BOOKKEEPER} \\ &\sf{3.APPEARING} \end{cases}\end{gathered}\end{gathered}$

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$\large\underline\blue{\bold{Formula \: used:- }}$

A permutation is the choice of r things from a set of n things without replacement and where the order matters.

$\boxed{\bf \:^{n}P_r=\dfrac{n!}{(n-r)!}}$

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$\large\underline\purple{\bold{Solution :- }}$

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$\bf \:\large \red{AηsωeR : 1.} ✍$

☆The word is STATISTICS.

☆In this word, there are 10 alphabets, in which 3 S's, 3 T's and 2 I's are there. So,

$\sf Number\:of\:ways =\dfrac{\:^{10 } P_{10}}{\:^{3} P_3\:^{3} P_3\:^{2} P_2}$

$\sf \:  = \dfrac{10!}{3! \times 3! \times 2!}$

$\sf \:  = \dfrac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3!}{3! \times 3 \times 2 \times 1 \times 2 \times 1}$

$\sf \:  = 50400$

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$\bf \:\large \red{AηsωeR : 2.} ✍$

☆The word is BOOKKEEPER.

☆In this word, there are 10 alphabets, in which 3 E's, 2 O's and 2 K's are there. So,

$\sf Number\:of\:ways =\dfrac{\:^{10 } P_{10}}{\:^{3} P_3\:^{2} P_2\:^{2} P_2}$

$\sf \:  = \dfrac{10!}{3! \times 2! \times 2!}$

$\sf \:  = \dfrac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3!}{3! \times 2 \times 1 \times 2 \times 1}$

$\sf \:  = 151200$

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$\bf \:\large \red{AηsωeR : 3.} ✍$

☆The word is APPEARING.

☆In this word, there are 9 alphabets, in which 2 P'sand 2 A's are there. So,

$\sf Number\:of\:ways =\dfrac{\:^{9} P_{9}}{\:^{2} P_2\:^{2} P_2}$

$\sf \:  = \dfrac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1 \times 2 \times 1}$

$\sf \:  = 90720$

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