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1. If
2. If
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 appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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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.
Always satisfy the Prime Directive of getting the right answer above all else.
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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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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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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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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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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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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.
Always satisfy the Prime Directive of getting the right answer above all else.
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