Question
Question: There are \(4n\) things out of which \(n\) are alike and the rest are different. Then the number of ...
There are 4n things out of which n are alike and the rest are different. Then the number of permutations of 4n things taken is 2n at a time each permutation containing n alike things.
option A: (n!)2(4n)!
option B: (2n)!(4n)!
option C: n!(4n)!
option D: (n!)2(3n)!
Solution
When there are n things, we have to select r things, the total number of ways for selecting is: r!(n−r)!n!
When there are n things and m similar things and we have to divide by the factorial of the number of similar things we are picking out and arranging.
These are very important for solving this question.
Complete step-by-step answer:
We are given that: there are 4n things out of this n are alike and the rest are different. Then the number of permutations of 4n things taken is 2n at a time each permutation containing n alike things.
Number of ways to pick n different things from 3n things is: (2n!)×(n!)(3n)!
Number of ways to pick n similar things from n alike things is: 1
Now, we have to find the number of ways of arranging these things. We have n similar things and n different things.
Total number of possibilities = (2n!)×(n!)(3n)!×1
Since, there are n similar things in 2n things, we have to multiply it with: n!(2n)!
The total number of ways will be: (2n!)×(n!)(3n)!×n!(2n)!
Now, simplifying the above equation we get: (2n!)×(n!)(3n)!×n!(2n)!=(n!)2(3n)!
So, the correct answer is “Option D”.
Note: To solve this question one needs to remember the formulae of permutations and combinations when there are similar things and different things. We have to divide with the factorial of the number of similar things we are arranging, here in this case it is n . Students make mistakes of not reading the question properly, one simple misconception will give you a completely wrong answer.