if the first 4 goes at location 1, the second 4 can go at 5 different locations, number of ways = 5 2, the second 4 can go at 4 different locations, number of ways = 4 3, the second 4 can go at 3 different locations, number of ways = 3 4, the second 4 can go at 2 different locations, number of ways = 2 5, the second 4 can go at 1 different location, number of ways = 1
Total number of different 6 digit numbers = 5+4+3+2+1 = 15
Option D
Good question BTW .....
Can you explain the bold part?I believe only 4 locations are available for the next four.
here with the three points its became equilateral triangle
so ans is 25sqrt3/4
but spherical ball so i think ans is less than 25sqrt3/4 (i am not sure")
Answer should be D, since body formed by tying 3 nails is not a 2 dimensional figure. equilateral triangle will be a projection of this body, so area has to be greater than 25 (sqrt 3)/4
Answer should be D, since body formed by tying 3 nails is not a 2 dimensional figure. equilateral triangle will be a projection of this body, so area has to be greater than 25 (sqrt 3)/4
I do not quite understand. How can a three dimesional equilateral triangle have greater area than that of two dimensional eq. triangle?
I think I was wrong. area should be same as that of equilateral triangle.
consider that you cut an equilateral triangle from a paper and paste it on a sphere, area still remains same.
D is the answer given and that is the reason I asked for your explanation. Your explanation matches some what with the answer explanation for this question. Here is the explanation given:
Take a piece of paper in the shape of an equilateral triangle and side 5cm and try to stick it in the spherical surface and try to cover the curved area marked by the strings completely. We will observe that the paper gets "ripped off" in between. This shows that we are trying to fit less area in larger surface. Hence, the (curved) area covered by the strings is more than the plain are of the figure formed by the same strings, which is in accordance with the equilateral triangle.
Answer is 5C2 which is 5!/(5-2)! 2! = 5!/3! 2! = 10.
_ 2 _ 0 _ 0 _ 2 _
1st 4 can be placed at any of the 5 positions. Once 1st 4 is placed, the 2nd 4 can be placed at any of the remaining 4 positions. So total number of ways 2 4's can be placed here is 5*4. Since both the number are 4 so positions have duplicate and the correct answer is 5*4/2 = 20/2 = 10.
no sir, your solution is wrong .... its not that simple a problem.... to prove my point .. here are 15 different numbers...
D is the answer given and that is the reason I asked for your explanation. Your explanation matches some what with the answer explanation for this question. Here is the explanation given:
Take a piece of paper in the shape of an equilateral triangle and side 5cm and try to stick it in the spherical surface and try to cover the curved area marked by the strings completely. We will observe that the paper gets "ripped off" in between. This shows that we are trying to fit less area in larger surface. Hence, the (curved) area covered by the strings is more than the plain are of the figure formed by the same strings, which is in accordance with the equilateral triangle.
in that case the answer should be "e) greater than or equal to 25(root 3)/4"
becuase no where the size of spherical ball is mentioned... so if i take a large ball ... say earth ... and make a 5 cm triangle ... what will be the area 😃
Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a second card is drawn. If the cards are drawn at random and if the sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is numbered 5 ?
Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a second card is drawn. If the cards are drawn at random and if the sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is numbered 5 ?
pls expln
combinations for sum 8 = (2,6) (6,2) (3,5) (5,3) (4,4).
the answer is B... you were close... all the values of X are in set (-1,0) U (1, infinite) and for all these values ... X > -1 will always be true...
if we select A, then we are leaving negative values of X..
Remember, the question is not asking all the possible values of X .. the question is "which of the following must be true for all X" ...
Only option B will always be true for all X
Do you want to say that X will belongs to the said group that's y it should be>-1? I believe that X>-1 will be considered as the domain and this has to be true for positive fractional value less than 1 too, which is definately not the case.
Do you want to say that X will belongs to the said group that's y it should be>-1? I believe that X>-1 will be considered as the domain and this has to be true for positive fractional value less than 1 too, which is definately not the case.
we dont have to find the domain for X ... we just need to find a condition that will be true for all X... our x is defined in first line....
Let me give you an example .. x is a member of set S = which of the following must be true for all x A. x >= 4 B. x >= -4
this is similar to the our original question... and the answer should be B
A contractor estimated that his 10-men crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, after 5 days of rain, on the 61-st day, he hired 6 more people to finish the project early. If the job was done in 100 days, on how many days after day 60 was it raining?
A contractor estimated that his 10-men crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, after 5 days of rain, on the 61-st day, he hired 6 more people to finish the project early. If the job was done in 100 days, on how many days after day 60 was it raining?