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So I have five different elements (A,B,C,D,E) and 2 spots they can fill. The question is how many unique pairs can I make, where the elements in the spots can be the same (non-unique). For this simple case the possible combos are (AA)(AB)(AC)(AD)(AE)(BB)(BC)(BD)(BE)(CC)(CD)(CE)(DD)(DE)(EE) or 5+4+3+2+1 = 15 different unique combos allowing repeated elements. My question is how can you find the answer for n elements and s spots? I'm sure there is a name for this kind of combination, but I don't know what it is. Am I missing a clever trick or a way to manipulate permutations and combinations which don't allow repeating elements? 24.106.205.226 02:43, 29 January 2007 (UTC)
Can
be rewritten without the summation into something that can be evaluated in a single spreadsheet cell? Neon Merlin 05:37, 29 January 2007 (UTC)
Is it true that the expected values of a sorted sample, size N, are the 1st, 3rd, 5th, ... and (2N-1)th 2N-quantiles? Neon Merlin 07:35, 29 January 2007 (UTC)
good aftenoon.... may i please know about the different mathematical formulas in solving problems which involved more than 7 digits? like for example finding the square root of a 7 digit figure in just a couple of second by just using the speed of my mental computation? thank you very much...your kind answer to this question would be a priceless help for my profession.... —The preceding unsigned comment was added by Never17fade ( talk • contribs).
To elaborate on a question asked a few days ago, I have a formula to determine the length of the angle bisectors, from vertex to the opposite side, in terms of the sides of the triangle. The length of the bisector from vertex A is 2*b*c*cos(A/2)/(b+c), the others having an obvious cyclic symmetry. cos(A/2) can be found in terms of a, b and c, of course.
But can it be done the other way round - is there an expression to get each side of the triangle from given values of the angle bisectors?… 86.132.237.140 21:07, 29 January 2007 (UTC)
My question, "can it be done ... " was shorthand for "please show me, as it looks rather non-invertible to me and I'd appreciate having the answer". I wasn't the asker of the numerical question a few days ago, by the way. If by side bisectors you mean medians, yes the sets of sides and medians are easy to get from each other.… 81.153.220.25 16:33, 30 January 2007 (UTC)
OK try again - I've derived (C2-a2)b = (C2-b2)a where C is the angle bisector length of the line that crosses triangle side c of length c.. The rest can be got by "obvious cyclic symmetry" - this looks simpler. For now I leave the rest to you. 87.102.13.207 17:39, 30 January 2007 (UTC)Apologies for not having been of any use. 87.102.13.207 18:19, 30 January 2007 (UTC)
Found it (missing square) (IGNORE the following)
cA = C2 + b2 -2Cbcos (angle C) and cB = C2 + a2 -2Cacos (angle C) therefore (cA - C2 - b2)/ -2Cb = cos (angleC) = ( cB - C2 - a2)/ -2Ca (cA - C2 - b2)/b = ( cB - C2 - a2)/a cAa - C2a - b2a = cBb - C2b - a2b (equation 1) but cB / cA = a/b see Angle bisector theorem so cB = cAa/b and cB + cA = c so cA + cAa/b =c cA (1 + a/b) =c cA ( (b+a)/b ) =c so cA =cb/(b+a) and cB =ca/(b+a) (equations 2) Combining equation1 and two equations 2 gives cba/(b+a) - C2a - b2a = cab/(b+a) - C2b - a2b C2a + b2a = C2b + a2b
I can't for the life of me see were I've gone wrong but I am getting tired so will come back tommorow. 87.102.13.207 19:41, 30 January 2007 (UTC)
Ignore below - made mistake..
C2 = ab So if I've not made a mistake (which is often the case) I get C2=ab B2=ac A2=bc 87.102.13.207 19:13, 30 January 2007 (UTC) So that gives the ratio of side lengths - another bit of info would probably finish off this problem - anyone? 87.102.13.207 19:22, 30 January 2007 (UTC)
cA2 = C2 + b2 -2Cbcos (angle C) and cB 2 = C2 + a2 -2Cacos (angle C)
therefore
(cA2 - C2 - b2)/ -2Cb = cos (angleC) = ( cB2 - C2 - a2)/ -2Ca
(cA2 - C2 - b2)/b = ( cB2 - C2 - a2)/a
cA2a - C2a - b2a = cB2b - C2b - a2b (equation 1)
but cB / cA = a/b see
Angle bisector theorem
so cB = cAa/b
and cB + cA = c
so cA + cAa/b =c
cA (1 + a/b) =c
cA ( (b+a)/b ) =c
so cA =cb/(b+a) and cB =ca/(b+a) (equations 2)
Combining equation1 and two equations 2 gives
(cb)2a/(b+a)2 - C2a - b2a = (ca)2b/(b+a)2 - C2b - a2b
Rearranging gives
abc = (C2-ab)(a+b)2/c and two other equations for A and B.. Not as simple.. SORRY. 87.102.13.207 20:08, 30 January 2007 (UTC)
(C2-ab)(a+b)2/c = (B2-ac)(a+c)2/b = (A2-bc)(c+b)2/a
Where capitals are bisector lengths, lower case are side lengths - solvable yes, easy no, multiple solutions don't know) 87.102.4.6 10:47, 2 February 2007 (UTC)
Mathematics desk | ||
---|---|---|
< January 28 | << Dec | January | Feb >> | January 30 > |
Welcome to the Wikipedia Mathematics Reference Desk Archives |
---|
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
So I have five different elements (A,B,C,D,E) and 2 spots they can fill. The question is how many unique pairs can I make, where the elements in the spots can be the same (non-unique). For this simple case the possible combos are (AA)(AB)(AC)(AD)(AE)(BB)(BC)(BD)(BE)(CC)(CD)(CE)(DD)(DE)(EE) or 5+4+3+2+1 = 15 different unique combos allowing repeated elements. My question is how can you find the answer for n elements and s spots? I'm sure there is a name for this kind of combination, but I don't know what it is. Am I missing a clever trick or a way to manipulate permutations and combinations which don't allow repeating elements? 24.106.205.226 02:43, 29 January 2007 (UTC)
Can
be rewritten without the summation into something that can be evaluated in a single spreadsheet cell? Neon Merlin 05:37, 29 January 2007 (UTC)
Is it true that the expected values of a sorted sample, size N, are the 1st, 3rd, 5th, ... and (2N-1)th 2N-quantiles? Neon Merlin 07:35, 29 January 2007 (UTC)
good aftenoon.... may i please know about the different mathematical formulas in solving problems which involved more than 7 digits? like for example finding the square root of a 7 digit figure in just a couple of second by just using the speed of my mental computation? thank you very much...your kind answer to this question would be a priceless help for my profession.... —The preceding unsigned comment was added by Never17fade ( talk • contribs).
To elaborate on a question asked a few days ago, I have a formula to determine the length of the angle bisectors, from vertex to the opposite side, in terms of the sides of the triangle. The length of the bisector from vertex A is 2*b*c*cos(A/2)/(b+c), the others having an obvious cyclic symmetry. cos(A/2) can be found in terms of a, b and c, of course.
But can it be done the other way round - is there an expression to get each side of the triangle from given values of the angle bisectors?… 86.132.237.140 21:07, 29 January 2007 (UTC)
My question, "can it be done ... " was shorthand for "please show me, as it looks rather non-invertible to me and I'd appreciate having the answer". I wasn't the asker of the numerical question a few days ago, by the way. If by side bisectors you mean medians, yes the sets of sides and medians are easy to get from each other.… 81.153.220.25 16:33, 30 January 2007 (UTC)
OK try again - I've derived (C2-a2)b = (C2-b2)a where C is the angle bisector length of the line that crosses triangle side c of length c.. The rest can be got by "obvious cyclic symmetry" - this looks simpler. For now I leave the rest to you. 87.102.13.207 17:39, 30 January 2007 (UTC)Apologies for not having been of any use. 87.102.13.207 18:19, 30 January 2007 (UTC)
Found it (missing square) (IGNORE the following)
cA = C2 + b2 -2Cbcos (angle C) and cB = C2 + a2 -2Cacos (angle C) therefore (cA - C2 - b2)/ -2Cb = cos (angleC) = ( cB - C2 - a2)/ -2Ca (cA - C2 - b2)/b = ( cB - C2 - a2)/a cAa - C2a - b2a = cBb - C2b - a2b (equation 1) but cB / cA = a/b see Angle bisector theorem so cB = cAa/b and cB + cA = c so cA + cAa/b =c cA (1 + a/b) =c cA ( (b+a)/b ) =c so cA =cb/(b+a) and cB =ca/(b+a) (equations 2) Combining equation1 and two equations 2 gives cba/(b+a) - C2a - b2a = cab/(b+a) - C2b - a2b C2a + b2a = C2b + a2b
I can't for the life of me see were I've gone wrong but I am getting tired so will come back tommorow. 87.102.13.207 19:41, 30 January 2007 (UTC)
Ignore below - made mistake..
C2 = ab So if I've not made a mistake (which is often the case) I get C2=ab B2=ac A2=bc 87.102.13.207 19:13, 30 January 2007 (UTC) So that gives the ratio of side lengths - another bit of info would probably finish off this problem - anyone? 87.102.13.207 19:22, 30 January 2007 (UTC)
cA2 = C2 + b2 -2Cbcos (angle C) and cB 2 = C2 + a2 -2Cacos (angle C)
therefore
(cA2 - C2 - b2)/ -2Cb = cos (angleC) = ( cB2 - C2 - a2)/ -2Ca
(cA2 - C2 - b2)/b = ( cB2 - C2 - a2)/a
cA2a - C2a - b2a = cB2b - C2b - a2b (equation 1)
but cB / cA = a/b see
Angle bisector theorem
so cB = cAa/b
and cB + cA = c
so cA + cAa/b =c
cA (1 + a/b) =c
cA ( (b+a)/b ) =c
so cA =cb/(b+a) and cB =ca/(b+a) (equations 2)
Combining equation1 and two equations 2 gives
(cb)2a/(b+a)2 - C2a - b2a = (ca)2b/(b+a)2 - C2b - a2b
Rearranging gives
abc = (C2-ab)(a+b)2/c and two other equations for A and B.. Not as simple.. SORRY. 87.102.13.207 20:08, 30 January 2007 (UTC)
(C2-ab)(a+b)2/c = (B2-ac)(a+c)2/b = (A2-bc)(c+b)2/a
Where capitals are bisector lengths, lower case are side lengths - solvable yes, easy no, multiple solutions don't know) 87.102.4.6 10:47, 2 February 2007 (UTC)