Q55 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Mathematics

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 $x+y$

  • A
    $\frac{10 !}{4 ! 2 ! 2 !}$
  • B
    $\frac{7 ! \times 15}{2 ! 2 ! 4 !}$
  • Correct answer
    C
    $\frac{6 !}{2 ! 2 !} \times 42$
  • D
    $\frac{7 !}{2 ! 2 !}+\frac{6 ! \cdot 7 p_4}{2 ! 2 !}$

Solution

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 $={ }^7 C_4 \times \frac{6 !}{2 ! 2 !}=y$ $\ldots(\mathrm{ii})$ $\therefore \quad x+y=\frac{7 !}{2 ! 2 !}+{ }^7 C_4 \times \frac{6 !}{2 ! 2 !}$ $=\frac{7 \times 6 !}{2 ! 2 !}+{ }^7 C_3 \times \frac{6 !}{2 ! 2 !}=\frac{6 !}{2 ! 2 !}\left(7+{ }^7 C_3\right)$ $=\frac{6 !}{2 ! 2 !}(7+35)=\frac{6 !}{2 ! 2 !}(42)$

Hence, the correct answer is option: 'C'.

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