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OP
rst

rst

Youngling
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NEW OBJECTIVE QUESTION

*img27.imageshack.us/img27/2843/penzr.png

Plz explain your answer

(Also you can ask your mathematics related questions or queries
we will try to solve them)
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Niilesh

Niilesh

Padawan
rst said:
total pen =11

So you use [sup]11[/sup]c[sub]4[/sub]

Then why you subract 1

plz explain ans in some more detail
Click to expand...
TBH i found that to be a little too easy that's why....

Let us assume that i take all 11 pens from both of them and then divide it into two groups of 4 & 7(pens are different). No. of ways dividing the the pens would be [sup]11[/sup]c[sub]4[/sub] = [sup]11[/sup]c[sub]7[/sub]. (we then give the new group of pens to both of them)
Now we subtract 1 because we also counted the initial configuration.
So ans is [sup]11[/sup]c[sub]4[/sub]
 
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OP
rst

rst

Youngling
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NEW OBJECTIVE QUESTION

*img850.imageshack.us/img850/7996/roser.png

Plz explain your answer

(Also you can ask your mathematics related questions or queries
we will try to solve them)
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Niilesh

Niilesh

Padawan
^ If we consider all roses of same colour to be identical then ans will be 1.

If we consider all of them distinct then my ans is coming 3600((6-1)! * [sup]5[/sup]c[sub]1[/sub] * 3!).
Can you give the answer in terms of factorial and/or C(n,r)?
 
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OP
rst

rst

Youngling
rst said:
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NEW OBJECTIVE QUESTION

*img850.imageshack.us/img850/7996/roser.png

Plz explain your answer

(Also you can ask your mathematics related questions or queries
we will try to solve them)
----------------------------------------------------------------------------------------------------------------------
Click to expand...
As white roses come together ,so we consider them as one flower
so we have 7 flowers
they can form garland in 6! ways
Also white flowers can be arranged in 3! ways
total ways =6! x 3!

As there is no difference between clockwise and anticlockwise permutation
So total ways =(6! x 3! )/2
=4320/2
=2160
 
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Niilesh

Niilesh

Padawan
rst said:
I think we have to consider all of them distinct
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As white roses come together ,so we consider them as one flower
Now total flowers = 7
they can form garland in 7! ways
Also white flowers can be arranged in 3! ways
total ways =7! x 3!
Click to expand...

They can be arranged in 6! in a garland not 7! as only order matter there is no positions in a circle. I also did it first by same method but i think some cases will be repeated in it.
 
Niilesh

Niilesh

Padawan
Point out mistake in this method

First i made a garland of red flowers in 5! ways. Then i selected 1 gap out of 5 gaps between the red flowers to put white flowers then arranged those flowers in 3! ways
 
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OP
rst

rst

Youngling
Niilesh said:
Point out mistake in this method

First i made a garland of red flowers in 5! ways. Then i selected 1 gap out of 5 gaps between the red flowers to put white flowers then arranged those flowers in 3! ways
Click to expand...

your method is correct
It will also give same result
 
OP
rst

rst

Youngling
hari1 said:
2^(x)=0
Find x. This is surely going to scratch your brains if you don't read it properly.
Click to expand...

Range of 2[sup]x[/sup] function is (0,∞)
So 2[sup]x[/sup]=0 is not possible for any real value of x
Hence 2[sup]x[/sup]=0 has no solution

In other words, there is no such x in R
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However, It is possible in extended real line i.e [-∞,∞]
2[sup]x[/sup]=0 is possible for x= -∞
 
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OP
rst

rst

Youngling
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NEW OBJECTIVE QUESTION

*img15.imageshack.us/img15/2318/probability.png

Plz explain your answer

(Also you can ask your mathematics related questions or queries
we will try to solve them)
----------------------------------------------------------------------------------------------------------------------
 
Niilesh

Niilesh

Padawan
^ No. of equilateral triangle possible = 6/3 = 2(6 vertices, 3 vertices have the same triangle)
Total No. of triangles possible 6C3 = 5*4 = 20
So probability = 2/20 = 1/10
 
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