Combinatorics DPP - RACE 3 - Grouping and Circular permutations easy

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 in 2 possible ways and then distribute to rooms in 3! ways:
Case 1: 1,2,2 => {5!/(1!2!2!2!)}.3! = 90
Case 2: 1,1,3 => {5!/(1!2!2!2!)}.3! = 60
Total: 150

(ii) In how many ways can five people be arranged in three different rooms if no room must be empty and each room has 5 seats in a single row.
Now we have to build upon the previous question's answer.
Case 1: 1,2,2 => {5!/(1!2!2!2!)}.3! = 90
Then arrange in each room: (5C1*1!) * (5C2*2!) * (5C2*2!) = 2000
So: 90 * 2000 = 180,000 
Case 2: 1,1,3 => {5!/(1!2!2!2!)}.3! = 60
Then arrange in each room: (5C1*1!) * (5C1*1!) * (5C3*3!) = 1500
So: 60 * 1500 = 90,000
Final answer: 90,000 + 180,000 = 270,000

Q4 - Find number of ways in which Rohit can score 100 in 20 consecutive balls, if on every ball he hits either six or four or plays a dot ball?
Solution:
x(6s) + y(4s) + z(dots) = 20
6x + 4y = 100
3x + 2y = 50
x = (50-2y)/3
x,y,z are integers
So
50-2y = 0 mod 3
2 = 2y mod 3
1 = y mod 3
Now iterate over y:
y = 1, x = 16, z = 3
y = 4, x = 14, z = 2
y = 7, x = 12, z = 1
y = 10, x = 10, z = 0

So:
20!/16!3!1! + 20!/4!14!2! + 20!/7!12!1! + 20!/10!10!0!

Q5 - In how many ways 5 persons can sit at a round table if two particular persons do not sit together.
Solution:
Arrange 3 people: (3-1)!
Now 3 gaps are there, choose 2: 3C2.
Arrange these 2 people: 2!
2!.3C2.2! = 12

Q6. In how many ways 4 boys and 4 girls can sit around a round table if no two girls are together.
Solution:
Arrange boys: 3!
Now 4 gaps. Only 1 way to choose gaps.
4! to arrange girls.
Answer 3!.4! = 144

Q7. Eight persons including A and B are seated on a circular table. How many arrangements are possible if exactly two persons sit between A and B:
First put A and B.
Now divide 6 people into groups of 2,4: 6!/2!4!
Now arrange each group: 2!.4!
Now 2! ways to arrange these groups.
Total: 6!.2! = 1440

Q8. Number of ways in which 13 different toys can be distributed between 3 brothers in such a way that distribution among the 2 elder brothers is even and the youngest one receives one toy more, is:
Solution:
x + x + x + 1 = 13 => x = 4
Divide toys in groups of 4,4,5 => 13!/4!4!2!5!
5 toys can only be given to the last brother.
Rest 2 can be distributed in 2! ways.
So finally: 13!/4!4!5!

Q9. In an eleven storeyed building (Ground floor + ten floor), 9 people enter in a lift cabin from ground floor. It is known that they will leave the lift in groups of 2, 3 and 4 at different floors. The number of ways in which they can get down is:
Solution:
Grouping: 9!/2!3!4!
Choose 3 floors: 10C3
Distribute groups on floors: 3!
Multiply all: 10!9!/7!2!3!4! = 10!.9.8/2.6.24 = 10!/4

Q10. A gentleman arranges a party of m + n (m != n) friends to a dinner and places m at one round table T1 and n at another round table T2. If not all people shall have the same neighbor in any two arrangements, then the number of ways in which he can arrange the guests, is:
Solution:
Group friends: (m+n)!/m!n!
Arranging m,n on tables: (m-1)!(n-1)!
Now we have to divide each of the table's arrangements by 2.
Why?
Because neighbors will be same in clockwise/anti clockwise arrangements.
Finally: (m+n)!/4mn

Q11. The number of ways in which 5 flowers from 8 different flowers can be strung to form a garland, is:
Solution:
8C5.4!/2 = 672

Q12. The number of ways in which 6 boys and 3 girls can stand in a circle so that all girls are together, is:
Solution:
Treat all girls as one and arrange: (7-1)! = 6!
Arrange 3 girls: 3!
Total: 6!.3! = 4320

Q13. The number of ways in 8 different beads can be threaded on to a ring such that 4 particular beads are always together, is:
Treat 4 beads as one: (5-1)! = 4!
Arrange 4 together beads: 4!
Divide by 2 for garland/ring/necklace: /2!
 = 288

Q14. In a SAARC summit, 2 Indians, 3 Sri Lankan, 3 Nepalis and 4 Pakistanis are to be seated for a round table conference. Total number of ways in which these delegates can take their seats if delegates of same nationality sit together, is:
Solution:
(4-1)!.2!.3!.3!.4! = 2!.(3!^3).4!