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Comment on The Fundamental Counting Principle
Thank you. very helpful.
Good point about electric
Good point!
This is the same diagram we
Very interesting!
Very interesting!
Source: Princeton review
Sandy has a husband and 2 children. She takes at
least 1 member of her family to a diner for lunch
every day. The diner offers 10 lunch specials. If no
one orders the same thing, how many different orders
can Sandy's family make for lunch?
Correct answer choices are 45, 90 and 720
I tried to solve this through the fundamental counting principle. I got 90 and 720 correct by multiplying 10*9 and 10*9*8. I wonder how the question reaches the 45 answer? I understand this can be solved through a combination formula, but I would like to solve it through fundamental counting. Can you help?
This is a VERY poorly-worded
This is a VERY poorly-worded question.
To begin, what constitutes ONE "lunch order"?
For example, if Sandy orders pasta and her husband orders steak, is that considered different from Sandy ordering steak and her husband ordering pasta? Or is it just considered ONE order of steak and pasta?
The answer choice of 45 suggests that it's considered ONE order of steak and pasta.
However, the answer choice of 720 suggests that this is NOT considered one order (with 3 people).
I suggest that you ignore this question.
Cheers,
Brent
Hii, very helpful video, pls
I realized in this question: https://gre.myprepclub.com/forum/in-state-x-all-vehicle-license-plates-have-2-letters-from-8626.html , zero was considered but in this question, https://gre.myprepclub.com/forum/number-of-two-digit-positive-integers-for-which-the-unit-2249.html, zero could not be considered for the tenths position, could you kindly clarify...thank you Sir
Good question!
Good question!
Positive 2-digit integers range from 10 to 99 inclusive.
Numbers with 0 in the tens position (01, 02, 03,...09) are not considered 2-digit numbers since the 0 in the tens position is not required. For example, 05 is the same as 5, and 08 is the same as 8.
In the second question (https://gre.myprepclub.com/forum/number-of-two-digit-positive-integers-f...), we're dealing with 2-digit numbers, so this means we must consider only 10 to 99 inclusive.
In the first question (https://gre.myprepclub.com/forum/in-state-x-all-vehicle-license-plates-h...), the last three spaces of the license plates must be one-digit numbers. In this case, each space can be filled with any of the 10 digits (0,1,2,3,4,5,6,7,8 or 9)
Cheers,
Brent
Thank you so much for this
You're not alone; most people
You're not alone; most people struggle with counting questions.
Glad you like the site!
Cheers,
Brent
Regarding this Kahn Academy
Bruce and Krista are going to buy a new furniture set for their living room. They want to buy a couch, a coffee table, and a recliner. They have narrowed it down so that they are choosing between 4 couches, 5 coffee tables, and 9 recliners.
How many different furniture combinations are possible?
https://www.khanacademy.org/math/probability/probability-geometry/counting-permutations/e/fundamental-counting-principle
The question asked for combinations.
I thought there were 4x5x9 permutations, and to find the number of combinations I would have to divide by 3! (in order to eliminate double counting of arrangements which are the same except in a different order. For example, Green Couch, Red Table, Blue Recliner is the same as Blue Recliner, Red Table, Green Couch.)
I'm sure there is a relatively simple flaw in my thinking. Could you please let me know what it is, and how to avoid it for these types of problems.
Thanks.
Great question!
Great question!
Unfortunately, the word "combinations" here is misleading.
If we think about solving the question in STAGES, we can see that there's no duplication.
Stage 1: Choose a couch.
Stage 2: Choose a coffee table.
Stage 3: Choose a recliner.
If we look at it this way, we are always choosing a couch first, a coffee table second and a recliner third.
So, our strategy does now allow for duplication because we're choosing a DIFFERENT kind of furniture at each stage.
That is, if we structure our solution this way, we can't choose a blue recliner first, then a red table, and then a green couch
----------------------
Now compare this to a SEEMINGLY similar question.
Joe has 7 friends (A, B, C, D, E, F and G) want to invite 3 to his party.
Stage 1: Choose a friend to invite.
Stage 2: Choose a friend to invite.
Stage 3: Choose a friend to invite.
In this case, we're drawing from the SAME GROUP of friends at each stage.
So, in this solution, we'll have duplication.
Does that help?
Cheers,
Brent
Yes, that was very helpful.
Hello! I am very confused on
It asks for the number of odd subsets out of 35. Can you break this down for me please? Usually with subsets you work backwards and do 35*34*33 depending on how many are in the subset. Thanks
When using the Fundamental
When using the Fundamental Counting Principle (aka the "slot method"), we must convince ourselves what each value in our product represents.
For example, in my solution (https://gre.myprepclub.com/forum/sets-with-odd-number-of-elements-15901....), the first stage is to determine whether element1 is included in the subset. Since element1 can be in the subset or not in the subset, there are TWO different ways to complete the first stage. So, 2 is the first value in my product.
In your solution, (35*34*33...), what does each value represent?
For example, the first value suggests there are 35 different ways to complete the first stage.
So, what's the first stage?
Six children ABCDEF and are
For this question I get how we can use the glue method. But I wanted to know why I cant make Betsy the most restrictive case and say there are six ways to sit B and only 2 ways to sit E next to be so we would then have 6*2*4*3*2*1 and then we can take that product and subtract it from 6!. Hope this makes sense. Please let me know if you need any clarification
I agree that we can place B
I agree that we can place B in 6 ways. However, there are more than 2 ways to place E.
For example, if B sits in chair #1, then E can sit in chair #3, 4, 5 or 6
Similarly, if B sits in chair #2, then E can sit in chair #4, 5 or 6
You can continue with your approach. You'll just need to consider two cases:
case i: B sits on one of the end chairs (chair #1 or 6)
case ii: B sits on one of the non-end chairs (#2, 3, 4 or 5)
If you continue with that approach, you'll find that case i can be completed in 192 ways, and case ii can be completed in 288 ways, for a total of 480 ways.
https://gmatclub.com/forum/of
FOr this problem I ended up getting 36 three times but why do we have to subtract 1 from 108 to get 107?
Question link: https:/
Question link: https://gmatclub.com/forum/of-the-3-digit-positive-integers-greater-than...
I'm assuming that you examined three possible cases:
- Integers in the form: same-same-different (e.g., 772) - 36 outcomes
- Integers in the form: same-different-same(e.g., 858) - 36 outcomes
- Integers in the form: different-same-same (e.g., 900) - 36 outcomes
However, the last set of outcomes includes the integer 600, which is not allowed, because we're told the integers must be greater than 600.
So we must subtract that one outcome from our sum.
Does that help?
https://gre.myprepclub.com/forum
Why cant we do the MISSISSIPPI rule here , we would have EESSNN right?
Question link: https:/
Question link: https://gre.myprepclub.com/forum/there-are-six-different-models-that-are...
Good idea, but we apply the Mississippi rule when some of the objects to be arranged are identical. For example, the four i's in Mississippi are all identical.
In this case, there are two models from Europe, but those models are not identical. In fact, the question tells us that the six models are all different.