Question

How many different words with or without meaning can be formed from-

• A. $\left(\frac{9!}{5!\times 2!}\right),\left(\frac{4!}{2!}\times \frac{5!}{2!}\right)$
• B. $\left(\frac{9!}{4!\times 2!}\right),\left(\frac{4!}{4!}\times \frac{5!}{2!}\right)$
• C. $\left(\frac{9!}{4!\times 2!}\right),\left(\frac{4!}{4!}\times \frac{6!}{2!}\right)$
• D. $\left(\frac{9!}{3!\times 2!}\right),\left(\frac{4!}{4!}\times \frac{5!}{3!}\right)$

Answer 1

The correct option is B $\left(\frac{9!}{4!\times 2!}\right),\left(\frac{4!}{4!}\times \frac{5!}{2!}\right)$

1. Total wordrs in "ALLAHABAD"=9, here "L" is repeated twice and "A" is repeated 4 times so total number of different words=$\left(\frac{9!}{4!\times 2!}\right)$
2. Those words in which vovels occupy even places$\to \frac{}{\mathrm{odd}}\frac{\left(\mathrm{vovel}\right)}{\mathrm{even}}\frac{}{\mathrm{odd}}\frac{\left(\mathrm{vovel}\right)}{\mathrm{even}}\frac{}{\mathrm{odd}}\frac{\left(\mathrm{vovel}\right)}{\mathrm{even}}\frac{}{\mathrm{odd}}\frac{\left(\mathrm{vovel}\right)}{\mathrm{even}}\frac{}{\mathrm{odd}}$total nuber of places for vovel=4$\to \mathrm{total}\mathrm{permutation}=\left(\frac{4!}{4!}\right),\mathrm{because}A\mathrm{is}\mathrm{repeated}4\mathrm{times}\mathrm{rest}\mathrm{places}\mathrm{remaining}\to 5\mathrm{now}\mathrm{remaining}\mathrm{permutation}\mathrm{of}\mathrm{words}\to \left(\frac{5!}{2!}\right),\mathrm{because}L\mathrm{is}\mathrm{repeated}\mathrm{twice}\mathrm{so}\mathrm{total}\mathrm{permutation}\to \left(\frac{4!}{4!}\times \frac{5!}{2!}\right)$option(b) is correct