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If $x$ and $y$ represent the number of arrangements of the letters of word ATRAPATRAM such that (i) all A's are together and (ii) no two A's are together respectively, then $x+y$
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The correct answer is:
$\frac{6 !}{2 ! 2 !} \times 42$
Given, word ATRAPATRAM
Here, $\mathrm{A} \rightarrow 4$ times
$\mathrm{T} \rightarrow 2$ times
$\mathrm{R} \rightarrow 2$ times
$\mathrm{P} \rightarrow 1$ time
$M \rightarrow 1$ time
(i) All A's are together,
Then it will be consider as one letter and remaining 6 letters and l'A's (including 4 A's) will be consider as 7 letters.
$\therefore \text { Number of arrangement }=\frac{7 !}{2 ! 2 !}=x$ $\ldots(\mathrm{i})$
(ii) no two $\mathrm{A}^{\prime}$ s are together.
such that $-\mathrm{T}-\mathrm{R}-\mathrm{P}-\mathrm{T}-\mathrm{R}-\mathrm{M}-$
$\therefore$ Number of arrangement $={ }^7 C_4 \times \frac{6 !}{2 ! 2 !}=y$ $\ldots(\mathrm{ii})$
$\therefore \quad x+y=\frac{7 !}{2 ! 2 !}+{ }^7 C_4 \times \frac{6 !}{2 ! 2 !}$
$=\frac{7 \times 6 !}{2 ! 2 !}+{ }^7 C_3 \times \frac{6 !}{2 ! 2 !}=\frac{6 !}{2 ! 2 !}\left(7+{ }^7 C_3\right)$
$=\frac{6 !}{2 ! 2 !}(7+35)=\frac{6 !}{2 ! 2 !}(42)$
Here, $\mathrm{A} \rightarrow 4$ times
$\mathrm{T} \rightarrow 2$ times
$\mathrm{R} \rightarrow 2$ times
$\mathrm{P} \rightarrow 1$ time
$M \rightarrow 1$ time
(i) All A's are together,
Then it will be consider as one letter and remaining 6 letters and l'A's (including 4 A's) will be consider as 7 letters.
$\therefore \text { Number of arrangement }=\frac{7 !}{2 ! 2 !}=x$ $\ldots(\mathrm{i})$
(ii) no two $\mathrm{A}^{\prime}$ s are together.
such that $-\mathrm{T}-\mathrm{R}-\mathrm{P}-\mathrm{T}-\mathrm{R}-\mathrm{M}-$
$\therefore$ Number of arrangement $={ }^7 C_4 \times \frac{6 !}{2 ! 2 !}=y$ $\ldots(\mathrm{ii})$
$\therefore \quad x+y=\frac{7 !}{2 ! 2 !}+{ }^7 C_4 \times \frac{6 !}{2 ! 2 !}$
$=\frac{7 \times 6 !}{2 ! 2 !}+{ }^7 C_3 \times \frac{6 !}{2 ! 2 !}=\frac{6 !}{2 ! 2 !}\left(7+{ }^7 C_3\right)$
$=\frac{6 !}{2 ! 2 !}(7+35)=\frac{6 !}{2 ! 2 !}(42)$
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