Guys PM's can be used for usual chit-chat.Find the sum of areas of all possible triangles that can be formed by joining the vertices of a cube of dimension 1 unit.
Consider a face of the cube, we can have 4 triangles by connecting the vertices.
Area of each of the 4 triangles -> 1/2
Total area of triangles that lie on a single face of the cube = 6 * 4 * 1/2 = 12 unit^2
Consider a single edge of the cube. Let this be 1 side of a triangle, then 2 triangles can be formed using 2 vertices of the diagonally opposite edge.
Area of each of the 2 triangles -> 1/2 * 1 * _/2
Total area of such triangles = 12 * 2 * 1/2 *_/2 = 12_/2
Consider traingles with vertices such that no 2 lie on the same edge. These triangles are basically triangles with sides that are otherwise diagonals of the faces of the cube.
If u take 1 such face diagonal, u can find 2 triangles of area _/3/4 (_/2)^2 = _/3/2 unit^2
So for 1 face with 2 diagonals, we have 4 triangles.
Total such triangles = 6 * 4 / 3 = 8 triangles.
Total area for such triangles = 8 * _/3/2 = 4_/3 unit^2
Total area = (12 + 12_/2 + 4_/3) unit^2 = 35.8 unit^2
A bug starts at one vertex of a cube and moves along the edges of the cube according to the following rule. At each vertex the bug will choose to travel along one of the three edges emanating from that vertex. Each edge has equal probability of being chosen, and all choices are independent. What is the probability that after seven moves the bug will have visited every vertex exactly once? (A) 1/2187(B) 1/729(C) 2/243(D) 1/81(E) 5/243
den i might be wrong 😞 ... thing is that there are only 3 ways in which a bug can reach covering all 7 vertices, and in 3^7 ways the bug can move... so 3/3^7 .. this seems easy... shouldn b so easy
The bug can move from its arbitrary starting vertex to a neighboring vertex in 3 ways. After this, the bug can move to a new neighbor in 2 ways (it cannot return to the first vertex). The total number of paths (as stated above) is 3^7 or 2187 . Therefore, the probability of the bug following a good path is equal to 6.x /2187 for some positive integer x . The only answer choice which can be expressed in this form is C.
The bug can move from its arbitrary starting vertex to a neighboring vertex in 3 ways. After this, the bug can move to a new neighbor in 2 ways (it cannot return to the first vertex). The total number of paths (as stated above) is 3^7 or 2187 . Therefore, the probability of the bug following a good path is equal to 6.x /2187 for some positive integer x . The only answer choice which can be expressed in this form is C. Exactly.
The bug can move from its arbitrary starting vertex to a neighboring vertex in 3 ways. After this, the bug can move to a new neighbor in 2 ways (it cannot return to the first vertex). The total number of paths (as stated above) is 3^7 or 2187 . Therefore, the probability of the bug following a good path is equal to 6.x /2187 for some positive integer x . The only answer choice which can be expressed in this form is C. Exactly.
Guys PM's can be used for usual chit-chat.Find the sum of areas of all possible triangles that can be formed by joining the vertices of a cube of dimension 1 unit.
So for 1 face with 2 diagonals, we have 4 triangles. Total such triangles = 6 * 4 / 3 = 8 triangles.Total area for such triangles = 8 * _/3/2 = 4_/3 unit^2Total area = (12 + 12_/2 + 4_/3) unit^2 = 35.8 unit^2
Now for 2nd partlet us assume A = a+6So, A + 6+b' + c'+d' =16Then,A+b'+c'+d' = 10So number of solutions = 13 c 3.
Take a closer look at the highlighted equation above. You could still have undesired cases where along with A, any of b', c', d' take values greater their required limits.
hi Can some one tell me whats wrong in my approach to the following question??Q. 4 dice are thrown.In how many ways can a sum of 20 be obtained?MY APPROACH ->we have a + b +c + d =20we have 2 conditions to be taken care of :->a,b, c and d >0 .....1>and a ,b,c,d So starting by 1>we have a'+1 +b' +1 +c' +1+d' +1 =20=>a'+b'+c'+d'=16So total solutions 19c3.Now for 2nd partlet us assume A = a+6So, A + 6+b' + c'+d' =16Then,A+b'+c'+d' = 10So number of solutions = 13 c 3.Now we can select any of A, b',c' and d' to be 6So final ans -> 19 c 3 - 4 *13 c 3But Problem is answer comes out to be -ve.Any help pointing out my mistake will be appreciated
What happens if you do 19C3 - 4 * 13C3 + 6 *7C3? (Comes to 35 I guess...OA is what?)
See, you need to also consider the cases where 2 dice cross 6. You have counted those cases twice and subtracted them, so need to add them back once. Venn diagram logic. regards scrabbler
hi Can some one tell me whats wrong in my approach to the following question??Q. 4 dice are thrown.In how many ways can a sum of 20 be obtained?MY APPROACH ->we have a + b +c + d =20we have 2 conditions to be taken care of :->a,b, c and d >0 .....1>and a ,b,c,d So starting by 1>we have a'+1 +b' +1 +c' +1+d' +1 =20=>a'+b'+c'+d'=16So total solutions 19c3.Now for 2nd partlet us assume a' = A+6So, A + 6+b' + c'+d' =16Then,A+b'+c'+d' = 10So number of solutions = 13 c 3.Now we can select any of A, b',c' and d' to be 6So final ans -> 19 c 3 - 4 *13 c 3But Problem is answer comes out to be -ve.Any help pointing out my mistake will be appreciated
I think in your second part..
you gotto take a' = a+7 instead of a+6 ... as you are taking cases where you need to take values more than 6...
hi Can some one tell me whats wrong in my approach to the following question??Q. 4 dice are thrown.In how many ways can a sum of 20 be obtained?
Alternative approach (oral solution alert @iLoveTorres)
Since the maximum possible score on the 4 dice is 24, the total shortfall for the 4 dice must be 4. So let the shortfalls be a, b, c, d and we have a + b + c + d = 4 hence 7C3 = 35 ways. regards scrabbler
Alternative approach (oral solution alert @iLoveTorres)Since the maximum possible score on the 4 dice is 24, the total shortfall for the 4 dice must be 4. So let the shortfalls be a, b, c, d and we have a + b + c + d = 4 hence 7C3 = 35 ways.regardsscrabbler
Exactly.