Question

The letter of word ALLAHABAD are arranged at random.the Probability that in word so formed,all similar letters are found together

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{word}"\mathrm{ALLAHABAD}"\mathrm{has}4\mathrm{A}\text{'}\mathrm{s},2\mathrm{L}\text{'}\mathrm{s},3\mathrm{distinct}\mathrm{alphabets}\left(\mathrm{HBD}\right)\mathrm{i}.\mathrm{e}\mathrm{total}\mathrm{of}9\mathrm{alphabets} \\ \mathrm{Total}\mathrm{cases}=\frac{9!}{4!2!}[4!\&2!\mathrm{are}\mathrm{divided}\mathrm{because}\mathrm{there}\mathrm{are}4\mathrm{As}\&2\mathrm{Ls}] \\ \mathrm{For}\mathrm{favourable}\mathrm{cases},\mathrm{we}\mathrm{combine}4\mathrm{A}\text{'}\mathrm{s}\mathrm{together}\&2\mathrm{L}\text{'}\mathrm{s}\mathrm{together},\mathrm{so},\mathrm{we}\mathrm{have}5\mathrm{alphabets}. \\ \mathrm{so},\mathrm{favourable}\mathrm{cases}=5! \\ \mathrm{required}\mathrm{probability}=\frac{\mathrm{favourable}\mathrm{cases}}{\mathrm{Total}\mathrm{cases}}=\frac{5!}{{\displaystyle \frac{9!}{4!.2!}}}=\frac{5!.4!.2!}{9!}=\frac{2\times 3\times 4\times 2}{6\times 7\times 8\times 9} \\ =\frac{1}{7\times 9}=\frac{1}{63} \end{gathered}$$