Question

In how many ways can $10$ lion and $6$ tigers be arranged in a row so that no two tigers are together?

• A. $10!\times {}^{11}{P}_6$
• B. $10!\times {}^{10}{P}_6$
• C. $6!\times {}^{10}{P}_7$
• D. $6!\times {}^{10}P$

Answer 1

Answer: (A)

There are $10$ lions and there is no restrictions on arranging lions. They can be arranged in $10!$ ways. But there is a restriction in arrangements of tigers that no two tigers come together. So two tiger are to be arranged on the either side of a lion. This gives $11$ places for tigers and there are $6$ tigers. So, tigers can be arranged in ${}^{11}{P}_6$ ways.
So, total arrangemns are $10!\times {}^{11}{P}_6$