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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 21,063

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.It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 104,733

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.****If it ain't broke, fix it until it is.** **A number by itself is useful, but it is far more useful to know how accurate or certain that number is.**

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

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

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

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

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

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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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 104,733

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.****If it ain't broke, fix it until it is.** **A number by itself is useful, but it is far more useful to know how accurate or certain that number is.**

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