consider a polygon of n sides.find the number of triangles,none of whose sides is the side of the polygon. ďťż
@nole said:consider a polygon of n sides.find the number of triangles,none of whose sides is the side of the polygon.
n(n-4)
@anantn said:@19rsbin addition to that how did you come up with 8 cases, can you elaborate on that?
where on numerator?
The 8 cases in the numerator
answer is nc3-2n-n*(n-4)c1 . may be this expression converges to ur expression or may be not.but if u get the correct answer please do share the approach.thanks in advance 😃
@anantn said:The 8 cases in the numerator
basically the regular polygon will be an regular octagon. So , 8 vertices there are. U can start from any one of the 8 vertices and take direction accordingly. So, 8 cases.
@nole said:answer is nc3-2n-n*(n-4)c1 . may be this expression converges to ur expression or may be not.but if u get the correct answer please do share the approach.thanks in advance
1 thing is not clear . the triangle will be formed through diagonals right?
There's the catch, the regular polygon formed doesn't need to be a hexagon, it can be any regular polygon, like a rectangle even, as long as the number of sides is a multiple of 8
I got 12/8^8
Er * square not a rectangle
@nole said:answer is nc3-2n-n*(n-4)c1 . may be this expression converges to ur expression or may be not.but if u get the correct answer please do share the approach.thanks in advance
not coming.. i am mistaken..
@anantn said:Question 2:In a class, of 10 students, any 5 students are selected for student council, each of the students can choose to vote for the president of council, and can vote for any of the other 4 students, or with old his/her vote. The student with the most number of votes becomes the president. What is the probability that the number of votes won by the president is a perfect square?Guys, please share entire approach along with the answer.
is it 5C1 X 2/(10C1 x 2^5 ) = 1 / (2 ^5)
Required case 5C1 x 2 ( 1 & 4 are perfect squares )
Total Cases = 10 C1 X 2^ 5 ( Each of the Five selected guys can vote or not i.e 2^5 )
@anantn said:There's the catch, the regular polygon formed doesn't need to be a hexagon, it can be any regular polygon, like a rectangle even, as long as the number of sides is a multiple of 8 I got 12/8^8
rectangle is not regular .. two square is possible
@anantn said:There's the catch, the regular polygon formed doesn't need to be a hexagon, it can be any regular polygon, like a rectangle even, as long as the number of sides is a multiple of 8 I got 12/8^8
but ur ques says "forms a regular polygon"
aur hexagon to kabhi nehi hoga
aur hexagon to kabhi nehi hoga
@hari_bang said:76% of 76% of 76% of 76% of..................................of 75
anyone wll tell me sol:)
@hedonistajay said:shudnt it be 4/270725 ... we can select ace, king, queen, and jack of exactly one suit ... in 4 ways ..
whts wrng in my sol
fr select anyoneee card out of ths 4 cards 16/52
aftrrrrrrr tht 2one...12/51
likee diss
(16/52)*(12/51)*(8/50)*(4/49)
@hari_bang said:whts wrng in my solfr select anyoneee card out of ths 4 cards 16/52aftrrrrrrr tht 2one...12/51likee diss(16/52)*(12/51)*(8/50)*(4/49)
16/52
bcz
4 cards of each...
@hari_bang said:What will be the probability of having 4 selected cards ace, king, queen, and jack of exactly one suit?
4/27025? 4*4C4/52C4
@hari_bang said:76% of 76% of 76% of 76% of..................................of 75 anyone wll tell me sol
0.76^n*75
=> 0.76^n
=> nlog0.76
log75 = log(25*3) = 2log5+log3 = 2(1-0.301) + 0.477 = 1.875
log0.76 = log76 - 2 = log38 + log2 -2 = log 2+ log2 + log10 + log2 -2 = -0.097
(rounding off log 38 to log40 = log4+log10 = 2*log2+log10 )
=> n((-0.097)
=> n> 1.875/0.097
=> n > 19.32
so i assume n min can be 20 😃
lots of assumptions so might be a near miss 😃