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#1 2008-11-20 23:34:25

ganesh
Moderator
Registered: 2005-06-28
Posts: 14,438

Pre-University Mathematics

1. If


find n.

2. If


find n.

3. From 6 boys and 4 girls, 5 are to be selected for admission to a particular course. In how many ways can this this be done if there must be exactly 2 girls?

4. Find the number of diagonals in a polygon of a side. How many triangles can be made?

5. Find the number of arrangements in which 6 boys and 4 girls can be arranged in a line so that all the girls sit together and all the boys sit together.

6. A family of 4 brothers and 3 sisters are to be arranged for a photograph in one row. In how many ways can they be seated if all the sisters sit together.

7. If 6 times the number of permutations of n things taken 3 together is equal to 7 times the number of permutations of (n-1) things choses 3 at a time, find n.

8. In an election, a voter may vote for any number of candidates not greater than the number to be chosen. There are 7 candidates and 4 members are chosen. In how many ways can a person vote?

9. Using Binomial theorem, find the value of

.

10. Find the coefficient of

in the expansion of

.


Character is who you are when no one is looking.

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#2 2009-05-21 11:32:35

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,666

Re: Pre-University Mathematics

Hi ganesh;

  These are pretty old so I won't hide them.

#1



n=8

#2 ignoring the trivial 0 and 1:
n=5

#9
By the binomial theorem:

If we say x=100 and y= -1 and we plug these into the RHS of the above identity. We get;



#10

Last edited by bobbym (2009-05-23 13:46:04)


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#3 2009-08-16 08:20:09

rzaidan
Member
Registered: 2009-08-13
Posts: 59

Re: Pre-University Mathematics

Hi ganesh

problem no.  7 :

6(nC3)=7((n-1)C3)

6*n(n-1)(n-2)/(3*2*1) =  7 (n-1)(n-2)(n-3)/(3*2*1) so by cancellation ((Hint :  n ≥3))

6n  =  7(n-3)  or   6n = 7n - 21  so n=21

Best Regards

Riad Zaidan

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#4 2009-08-16 08:24:24

rzaidan
Member
Registered: 2009-08-13
Posts: 59

Re: Pre-University Mathematics

Hi ganesh

problem no.  3 :

The no. of ways = (6C3)(4C2)= 6(5)(4)/3(2)(1) *  4(3)/2(1)

                 =20*6=120


Best Regards

Riad Zaidan

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#5 2009-08-16 08:27:44

rzaidan
Member
Registered: 2009-08-13
Posts: 59

Re: Pre-University Mathematics

Hi ganesh

problem no.  4:

The no. of diagonals= a(a-3)/2
               

The no. of triangles= aC3=a(a-1)(a-2)/3*2*1

Best Regards

Riad Zaidan

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#6 2009-08-16 08:33:08

rzaidan
Member
Registered: 2009-08-13
Posts: 59

Re: Pre-University Mathematics

Hi ganesh

problem no.  5:

The number of arrangements= 2 * 6! * 4!
                     
                                           2*6(5)(4)(3)(2)(1)*4(3)(2)(1)

Notice that the 2 in the expression denotes the number of cases for the side can be chosen

(  left or right of the line)

Best Regards

Riad Zaidan

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#7 2009-08-16 08:39:06

rzaidan
Member
Registered: 2009-08-13
Posts: 59

Re: Pre-University Mathematics

Hi ganesh

problem no.  6:

First we choose the three neighbouring positions for the girls by 5 ways so

they can be seated by   5 * 4! * 3! =5(4)(3)(2)(1)*(3)(2)(1)=720 ways


Best Regards

Riad Zaidan

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#8 2009-08-16 09:15:25

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,666

Re: Pre-University Mathematics

Hi ganesh;

Last edited by bobbym (2009-08-16 09:20:33)


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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