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Comment on When to use Combinations
ausm! thnx!! :)
Amazing explanation. Crisp
Glad you like it!
Glad you like it!
Hi!
In the second question listed (urch webpage) I couldn't see the question even after scrolling to bottom. Could you look into that?
Thanks
Thanks for the heads up.
Thanks for the heads up.
That site has had several problems recently, so I have deleted the link.
Cheers and thanks,
Brent
Combinations have always been
So how do we determine there
You're referring to the
You're referring to the question that starts 3:55 in the above video.
I'm not sure what you're asking when you say "So how do we determine there are no different toppings"
Can you please elaborate?
Cheers,
Brent
Superb explanation.
the problem -
How many 3 -digit number greater than 399 ending eith 5 or 7?
Kindly let me know if I can proceed as below steps:-
Let us take there are no restriction of 5 or 7 so
First digit - 6 ways
Second digit - 10 ways
Third digit - 10 ways.
So total no. of ways = 6 * 10 * 10 =600
Now let us assume the last digit is not filled by 5 or 7
so,
First digit - 6 ways
Second digit - 10 ways
Third digit - 8 ways.
So total no. of ways = 6 * 10 * 8 =480 ways.
Therefore no. ways the number greater than 399 and the last digit is 5 or 7 = 600 -480 = 120 ways
You're referring to the
You're referring to the question that starts at 5:10 in the above video.
Yes, that approach works perfectly. It requires some extra steps, but the important thing is that you got the correct answer. Nice work!
Cheers,
Brent
Thanks Brent
Hello Brent,
Can you please clarify me, what difference in answer is actually observed when solving same question by fundamental counting principle as well as combination technique?
If you're referring to the
If you're referring to the question that starts at 5:10 in the above video, it can't be solved using the combinations.
Here's my solution:
The question: How many 3-digit integers numbers greater than 399 end with 5 or 7?
Stage 1: Select the last (units) digit.
This digit must be 5 or 7
So, the stage can be completed in 2 ways
Stage 2: Select the first (hundreds) digit.
This digit must be 4, 5, 6, 7, 8 or 9
So, the stage can be completed in 6 ways
Stage 3: Select the second (tens) digit.
This digit must be 0, 1, 2, 3, 4, 5, 6, 7, 8 or 9
So, the stage can be completed in 10 ways
By the fundamental counting principle, the total number of 3-digit number that satisfy the given conditions = (2)(6)(10) = 120
Cheers,
Brent
I can understand the logic
The fast way looks like this:
The fast way looks like this:
12C3 = (12)(11)(10)/(3)(2)(1) = 220
Two videos from this video, (at https://www.greenlighttestprep.com/module/gre-counting/video/789) you'll learn how I calculated 12C3 above.
Cheers,
Brent
You know what's interesting
Good idea!
Good idea!
For the handshake example, I
That's the beauty of listing
That's the beauty of listing and counting!
If you list possible outcomes in a systematic way, you'll often see a pattern (e.g., 7 + 6 + 5 ....)
Hello - in 6:22. Why did you
The middle (tens) digit can
The middle (tens) digit can be any digit (0,1,2,3,4,5,6,7,8 or 9)
So, there are 10 ways to complete that stage.
Hi Brent, it seems below page
Page not found
The requested page "/articles/combinations-and-non-combinations-%e2%80%93-part-iibinations-%e2%80%93-part-ii" could not be found.
Fixed! Thanks for the heads
Fixed! Thanks for the heads up.
Great thanks Brent.