.
= = cancel 3! in the numerator and denominator = = = 10. Was it hard ?
= = cancel 4! in the numerator and denominator = = 7*5 = 35. Was it hard ?
= = cancel 3! in the numerator and denominator = = 5*4 = 20. Was it hard ?
Now the final answer is 10*35*20.
Can you calculate it on your own ?
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On Combinations and Permutations see the lessons
- Introduction to Permutations
- PROOF of the formula on the number of Permutations
- Problems on Permutations
- Introduction to Combinations
- PROOF of the formula on the number of Combinations
- Problems on 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.
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Hello, Lisa,
I got your comment regarding my typo in the lesson "Introduction to permutations".
You are right. It was a typo. I just fixed it.
Thank you for your noticing !
Regarding number of combinations of "n" items taken "m" at a time, there is the formula
= .
(See my lessons related to Combinations).
You can always cancel one of the factors m! or (n-m)! in the numerator and denominator.
H a p p y l e a r n i n g ! !