@viewpt said:Q: in how many ways one can paint all the sides of a hexagon?
neither clarity hai Q mein..kitne colors and all..nor options 
@jain4444 said:number of squares ?
16
@viewpt said:Q: in how many ways one can paint all the sides of a hexagon?
all same->1 way
5 same , 1 diff ->6c1 * 2!
4 same ,2 diff -> 6c2*3!
4 same 2 same ->6c2*2!
3same 3 same ->6c3 * 2!
3 same 2 same 1 dif -> 6c3*3c2*3!
3 same 3 diff -> 6c3*4!
add all above?
@jain4444 said:all correct@viewpt pranam bhai With the use of three different weights, namely, 1gm, 3gm , 9gm and 27gm . In how many ways different weights can be weighed, if the objects to be weighed and the given weights may be placed in either pan of the scale ? (both pans are not identical)
both pans are nt identical means ?
...will update here....
@jain4444 said:all correct@viewpt pranam bhai With the use of three different weights, namely, 1gm, 3gm , 9gm and 27gm . In how many ways different weights can be weighed, if the objects to be weighed and the given weights may be placed in either pan of the scale ? (both pans are not identical)
3 powers soo
1+3+9+27 = 40 so all upto 40 ?
wt does this both pans nt identical mean?? 😲
@viewpt said:Q: in how many ways one can paint all the sides of a regular hexagon?
All sides can only be painted in 1 way, rather question should be "In how many ways one can paint the sides of a regular hexagon?"
Then answer would be 12, considering symmetry.
For painting 1 side - 1way
2-3Ways
3-3Ways
4-3Ways
5-1Ways
6-1Ways
Then answer would be 12, considering symmetry.
For painting 1 side - 1way
2-3Ways
3-3Ways
4-3Ways
5-1Ways
6-1Ways
@Brooklyn said:3 powers soo1+3+9+27 = 40 so all upto 40 ?
here non identical is given, confusion go rahi hai
@sujamait said:both pans are nt identical means ?...will update here....
@Brooklyn said:3 powers soo1+3+9+27 = 40 so all upto 40 ?wt does this both pans nt identical mean??
ek red color ka hai aur ek green 
@jain4444 said:all correct@viewpt pranam bhai With the use of three different weights, namely, 1gm, 3gm , 9gm and 27gm . In how many ways different weights can be weighed, if the objects to be weighed and the given weights may be placed in either pan of the scale ? (both pans are not identical)
40 tak kar sakte h.
3 ki power ki series h
to jo measure ho sakta h wo in sab ka sum hota h 1+3+9+27=40
ab ques k according 40 *2 =80 ways h answer
Tell me if i am wrong 😃
@jain4444 said:all correct@viewpt pranam bhai With the use of three different weights, namely, 1gm, 3gm , 9gm and 27gm . In how many ways different weights can be weighed, if the objects to be weighed and the given weights may be placed in either pan of the scale ? (both pans are not identical)
Another way of doing. Which is actually the same as the one mentioned by you a couple of pages earlier.
Every weight can be placed in 3 ways (Left pan/Right Pan/No Pan)
So total ways
= 3x3x3x3 -1 [Minus 1 for the case that no wieght is placed in any pan]
= 80
Every weight can be placed in 3 ways (Left pan/Right Pan/No Pan)
So total ways
= 3x3x3x3 -1 [Minus 1 for the case that no wieght is placed in any pan]
= 80
@sujamait said:both pans are nt identical means ?...will update here....
I guess it means that we can go from 40 to -40 i.e. 81 possibilities for the difference between the pans...but the number of distinct weights we could weigh would be 1 till 40 only.
Or another way to look at it is, we can put each weight in either pan or in neither so each of the weight has 3 possibilities so for 4 weights we have 3^4 = 81 possibilities (since the weights are powers of 3, there is no way any two can add up to give a third so there will be no overlaps)
regards
scrabbler
Or another way to look at it is, we can put each weight in either pan or in neither so each of the weight has 3 possibilities so for 4 weights we have 3^4 = 81 possibilities (since the weights are powers of 3, there is no way any two can add up to give a third so there will be no overlaps)
regards
scrabbler
@scrabbler said:I guess it means that we can go from 40 to -40 i.e. 81 possibilities for the difference between the pans...but the number of distinct weights we could weigh would be 1 till 40 only.Or another way to look at it is, we can put each weight in either pan or in neither so each of the weight has 3 possibilities so for 4 weights we have 3^4 = 81 possibilities (since the weights are powers of 3, there is no way any two can add up to give a third so there will be no overlaps)regardsscrabbler
@scrabbler said:I guess it means that we can go from 40 to -40 i.e. 81 possibilities for the difference between the pans...but the number of distinct weights we could weigh would be 1 till 40 only.Or another way to look at it is, we can put each weight in either pan or in neither so each of the weight has 3 possibilities so for 4 weights we have 3^4 = 81 possibilities (since the weights are powers of 3, there is no way any two can add up to give a third so there will be no overlaps)regardsscrabbler
ohk..yeah realized Q asks for "In how many ways different weights can be weighed"
@viewpt said:3^4=81hence till 80 then two diff pans toh 80*2=160 whr gettting wrong??
3^4 is when you are considering the two pans.
For every weight you have three possibilities
i) It is not used
ii) It is on Pan 1
iii) It is on Pan 2
Now, there are 4 weights, so 3^4 ways (but we need to remove the case when none of the weight is chosen)
So, 3^4 - 1 = 80 ways
Let N be the smallest positive integer with six positive odd integer divisors and twelve positive even integer divisors. Find the sum of the digits of N?
OPTIONS
1) 18
2) 9
3) 27
4) 15
OPTIONS
1) 18
2) 9
3) 27
4) 15
Let the roots of x^3 + 3x^2 + 4x – 11 = 0 be a, b, c and the roots of x^3 + rx^2 + sx + t = 0 are a + b, b + c, c + a the find the value of t?
OPTIONS
1) 0
2) –11
3) 23
4) 98
OPTIONS
1) 0
2) –11
3) 23
4) 98
thak gaye :|
@sujamait said:Let N be the smallest positive integer with six positive odd integer divisors and twelve positive even integer divisors. Find the sum of the digits of N?OPTIONS1) 18 2) 9 3) 27 4) 15
9?
The number will be 180 I think...2^2 x 3^2 x 5^1
regards
scrabbler
The number will be 180 I think...2^2 x 3^2 x 5^1
regards
scrabbler
@sujamait said:Let N be the smallest positive integer with six positive odd integer divisors and twelve positive even integer divisors. Find the sum of the digits of N?OPTIONS1) 18 2) 9 3) 27 4) 15
Total factors=18=2 * 3^2
(1+1)(2+1)(2+1)
Smallest will be 2^2*3^2*5=180
Sum 9?
(1+1)(2+1)(2+1)
Smallest will be 2^2*3^2*5=180
Sum 9?
@sujamait said:Let the roots of x^3 + 3x^2 + 4x €“ 11 = 0 be a, b, c and the roots of x^3 + rx^2 + sx + t = 0 are a + b, b + c, c + a the find the value of t?OPTIONS1) 0 2) €“11 3) 23 4) 98 thak gaye
23?
@sujamait said:Let the roots of x^3 + 3x^2 + 4x €“ 11 = 0 be a, b, c and the roots of x^3 + rx^2 + sx + t = 0 are a + b, b + c, c + a the find the value of t?OPTIONS1) 0 2) €“11 3) 23 4) 98 thak gaye
23 aa raha hai...just rechecking....
regards
scrabbler
regards
scrabbler