If there’s 18 spots open for a College class, and there’s 30 kids, how many ways can you fill those spots?
Is this a combination?
@ThatSpellingGuy111 the simplest explanation I can think of is:
Permutation: Picking ABCD is different from picking ABDC or CADB.
Combination: Order is irrelevant, so ABCD is equivalent to ABDC or CABD or any of the 4! = 24 rearrangements.
In general, the number of ways to select k objects from n, with order relevant, is n(n-1)(n-2)…(n-k+1) = n!/(n-k)! (sometimes denoted (n)_k) This is the number of k-permutations of n, or the number of sequences of k different objects selected from n.
The number of ways to select k objects from n, with order irrelevant is (n(n-1)(n-2)…(n-k+1))/k! = n!/((n-k)!k!). This is the number of combinations of k objects from a set of n objects, which is equal to the number of k-element subsets of {1,2,3,…,n}. It is usually denoted nCk or, in LaTeX, \binom{n}{k}.
Note that you have to divide by k! because each k-element set (when finding permutations) is counted k! times.
Yes. How many ways?
I discuss Permutations and Combinations here: http://talk.collegeconfidential.com/discussion/comment/15836111/#Comment_15836111
and here: http://talk.collegeconfidential.com/discussion/comment/15871006/#Comment_15871006
And here is my discussion on picking numbers: http://talk.collegeconfidential.com/discussion/comment/15771861/#Comment_15771861
@MITer94
So some keywords that might indicate permutations would be “distinct ways” and “how many different ways” can something be arranged?
@MITer94 I got 86493225 ways.
@ThatSpellingGuy111 Not really. What you want to look for is, “does the order of which I am picking my objects matter?” If the answer is yes, use permutations. Otherwise, combinations.
Yes. That is the value of 30C18 (or 30C12).
@DressingIron “The price of ground coffee beans is d dollars for 8 ounces and each makes c cups of brewed coffee. In terms o c and d , what in the dollar cost of the ground coffee beans required to make 1 cup of brewed coffee ?” Did you mean “each ounce makes c cups of brewed coffee”?
A necklace is formed by stringing 133 colored beads on a
thin wire in the following order: red, orange, yellow, green,
blue, indigo, violet; red, orange, yellow, green, blue, indigo,
violet. If this pattern continues, what will be the color of the
101st bead on the string?
(A) Orange (B) Yellow © Green (D) Blue (E) Indigo
If we divide 101 by 7 we get a remainder of 4 so the color should be green right but in the Barron’s book the answer is given B.Can someone please help me out.
@UAS998 The remainder is 3. Use your calculator.
Oh yeah!!! oops sorry wrong calculation,Thanks a lot for the help
By how many degrees does the angle formed by the
hour hand and the minute hand of a clock increase from
1:27 to 1:28?
Please help ASAP.
Thanks in Advance.
@UAS998 The minute hand moves clockwise by 360/60 = 6 deg. But the hour hand also moves clockwise, but only by 30/60 = 0.5 deg (30 degrees formed by 1:00 and 2:00, divided by 60 min). The angle changes by 6 - 0.5 = 5.5 deg.
Oh!
Thank’s a lot.
@MITer94 Hey, I have this math question that I know how to do by plugging in values, but how do I do it without plugging and chugging?
It is a number line that has these points in this order: A, C, M, B, E, D
The question is:
In the figure above, B is the midpoint of MD, M is the midpoint of AB, C is the midpoint of AM, and E is the midpoint of BD. The length of AM is what fraction of the length of AE?
What would you do algebraically?
@ThatSpellingGuy111 plugging in a length (e.g. AC = 1) is completely valid, since all ratios are preserved no matter what you plug in. To do it more generally, you could let AC = x and do the same thing.
http://imgur.com/VzXTYUa
Can anyone help me with this…
This is not a SAT problem but i’m having trouble with this question more than most of the SAT problems.
Angela is wrapping 1 meter of twine around a spool with a 2-centimeter diameter. The spool is thin and accommodates only one wrap of twine before the twine stacks on top of itself. The twine has a diameter of 1/2 cm.
A) Find how many complete times Angela will wrap the twine
around the spool.
B) Find the percentage of a complete circle that the last wrapping of the twine will make. Round to the nearest tenth.
The answer on the online worksheets seems to be A=6 and B=60.4% but I’ve been trying since an hour and I just can’t get to these answers 
Hoping for somebody to provide a good explanation but any help would be appreciated.
^ That is a basic geometry problem that requires a bit of reasoning.
The following table might help you understand how to solve it. Your worksheet should like this.
1 2 3.1415 6.283 6.283
2 3 3.1415 9.4245 15.7075
3 4 3.1415 12.566 28.2735
4 5 3.1415 15.7075 43.981
5 6 3.1415 18.849 62.83
6 7 3.1415 21.9905 84.8205
7 8 3.1415 25.132 109.9525 too much go back to 6 but keep length to complete a rotation of 8 cm.
[QUOTE=""]
15.1795 0.603990928 of 25.132
[/QUOTE]
Note that an alternative method is to eliminate Pi in a preliminary step. A bit shorter and your paper should like this
1 2 1 2
2 3 1 3 5
3 4 1 4 9
4 5 1 5 14
5 6 1 6 20
6 7 1 7 27
1
7 8 1 8 35
Max 31.83192742 4.831927423 0.603990928
Now try to figure how the above works …
@AurangFall17 First revolution takes roughly 2pi cm of twine (not exactly though). The next revolution takes roughly 3pi cm since the diameter is now 2cm + 2*(0.5cm). Since
2pi + 3pi + 4pi + … + 7pi ~ 84.8 (cm)
2pi + 3pi + 4pi + … + 8pi ~ 110 (cm) > 100,
we conclude that the twine makes six full revolutions. The leftover part is 15.2 cm, which is wrapped around a circle of circumference 8pi, and 15.6/8pi is about 60.4%. I don’t understand why this problem asks for nearest tenth, because this problem doesn’t take into account elasticity, shape of the twine when it’s revolving, etc.