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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 15,870

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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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,400

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.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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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: 92,400

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.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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