.
I will solve ONLY the case B here.
It is the same as to ask:
In how many ways 6 different pigeons can be placed in 10 pigeonholes, under the condition
that there are no two pigeons in one pigeonhole ?
The answer is: in 10*9*8*7*6*5 = 151200 ways.
1-st pigeon can be placed into any of 10 pigeonholes;
2-nd pigeon can be placed into any of remained 9 pigeonholes;
3-rd pigeon can be placed into any of remained 8 pigeonholes;
4-th pigeon can be placed into any of remained 7 pigeonholes;
5-th pigeon can be placed into any of remained 6 pigeonholes;
6-th pigeon can be placed into any of remained 5 pigeonholes.
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On Combinations, see the lessons
- Introduction to Combinations
- PROOF of the formula on the number of Combinations
- Problems on Combinations
- Arranging elements of sets containing indistinguishable elements
- Persons sitting around a circular table
- Combinatoric problems for entities other than permutations and combinations
- OVERVIEW of lessons on Permutations and Combinations
in this site.
Also, you have this free of charge online textbook in ALGEBRA-II in this site
- ALGEBRA-II - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this online textbook under the topic "Combinatorics: Combinations and permutations".
Save the link to this textbook together with its description
Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson
into your archive and use when it is needed.