Math IOQM

Required theory: Grouping of objects  and  Circular Permutations . Q1  - In how many ways 20 different books can be distributed between 7 students such that 3 students receive 4 books each and the remaining students receive 2 books each. Solution: First find the ways of grouping them in the group sizes of 4,4,4,2,2,2,2. 20!/(4!.4!.4!.3!).(2!.2!.2!.2!.4!) Then we need find ways of labeling them under 7 different labels. 7! Multiply both to get the final answer: 7!.20!/(4!.4!.4!.3!).(2!.2!.2!.2!.4!) Q2  - In how many ways 5 boys and 4 girls can sit in straight line if 2 girls are together and the other 2 girls are also together but separated from the first 2. Solution : First arrange 5 boys: 5! Now there are 6 gaps, choose 2: 6C2 Now if you see carefully, the 4 girls can be arranged in any order in these 2 gaps, so 4! Finally: 5!.6C2.4! = 43200 Q3  - (i) In how many ways five people can be distributed in three different rooms if no room must be empty? Make groups ...

How do we divide 7 distinct objects into groups of 3 and 4? It's quite simple. We just need to select 3 objects from 7. So it's 7C3. Which is same as 7!/(3!4!). Ok, now how do we create groups of 3 and 3 from 6 objects? Is it 6C3? No, there is an issue here. Let's say the objects are A,B,C,D,E,F. When we do 6C3, we will get (A,B,C), (D,E,F) as well as (D,E,F), (A,B,C). Which are essentially the same groups. So we will have to divide by 2!. 2 here is the number of groups of the same size. So if we have to divide (m + n) objects into groups of m and then: 1. if m != n then (m+n)!/(m!n!) 2. if m = n then (2m)!/(m!m!2!) ----------------------------------------------------------------- So far we have discussed formation of 2 groups. Now let's discuss more than 2 groups. ------------------------------------------------------------------ Let's say we have to divide 9 objects into groups of 2,3,4. Now first we choose 2 from 9 and then choose 3 from remaining 7. So it become...