[yogendra]

Originally Posted by Rishabh19

Hi all,

Can sum1 plz.. provide me solution fr dis question...

Q) 17 students of std 10 of St.Ivan's school are ranked on d basis of the marks scored in their 9th exam. It is known that each student got different marks. A class council of 7 students is to be formed such athat it has a mix of different types of students. So, the class teacher has decided not to include any two students having consecutive ranks. In how many ways can the council be selected?

1) 7920 2) 120 3)462 4) 330 5) None of these

Guys/ gals plz.... provide me an xplanation. dis prob has made me insomniac

is the answe 4) 330

[yogendra]

approach-->

if we reframe the question no. of ways in which 7 guys will never sit together out of 17 guys.. question becomes simple

10 guys will sit first that will create 11 spaces for other 7 guys

so rest 7 guys can be arranged in 11P7 ways

or coming to the question they can be selected in 11C7 ways

[MaskedMenace]

Originally Posted by Leonayas

3)9*10*10*10*4*4
5) 5c3 * 4C3 * 6!

Both are incorrect

[Rishabh19]

Yes d answer is 330.

IS der any other way to approach dis kinda problem..???

[Leonayas]

Originally Posted by Leonayas

3)9*10*10*10*4*4
5) 5c3 * 4C3 * 6!

Originally Posted by MaskedMenace

Both are incorrect

3)9*9*8*7*4*4
5)5^3 * 4^3 * 6!
is it ok now?

[yogendra]

Originally Posted by Rishabh19

Yes d answer is 330.

IS der any other way to approach dis kinda problem..???

Another way can..

Let the 7 guys be a1 a2... a7 sitted in same order from left to right in a row

Let L1 , L2 ... L7 be number of persons sitting on left of a1,a2 respectively..
and R7 be no. of person on right side of a7

so if add up all a's L's and R sum will be 17
or L1+L2+...L7+R=10

now point we have to see here is
L1 and R will be greater than or equal to zero

while L2,L3... L7 will be greater than or equal to 1

if we take Y2=L2-1 ; then Y2 >= 0

same ways equation will reduce to L1+Y2+Y3+...Y6+R = 10-6

so nw its a question of placing 4 balls in 8 boxes where each box can hold any no. of balls

which is (8+4-1)C(8-1)
or 11C7


Hope these equations are clear.. tough problem

This approach might be mathematical but its more time consuming and prone to error.

[MaskedMenace]

Originally Posted by Rishabh19

Yes d answer is 330.

IS der any other way to approach dis kinda problem..???

Consider there are 10 white sticks and 7 red sticks

place 10 white sticks in some order, say left to right

it creates 11 gaps in between them, place 7 red sticks in those gaps,

we can choose 7 gaps out of 11 to place 7 red sticks in 11c7 = 330 ways.

[MaskedMenace]

Originally Posted by Leonayas

3)9*9*8*7*4*4
5)5^3 * 4^3 * 6!
is it ok now?

you are ignoring the cases when zero appears..

for first question

consider four cases

1)no zeros
2)exactly 1 zero
3)exactly 2 zeros
4)exactly 3 zeros

for second question

zero is also even

[Durba]

This is a common Q, however i am stuck at the approach. Why 1 method works and the other doesnt.
Q > Find the number of triangles whose vertices are vertices of an octagon but none of the sides of such triangle are taken from the sides of the octagon.

My approach >
the Q can be visualized as the number of ways of choosing vertices so that no two are consecutive. If the vertices are a,b,c,d,e,f,g,h and i pick "a" then let there be X1 vertices between "a" and the next vertices. Similarly X2 be the number of vertices b/w 2nd & 3rd vertices and X3 be number of vertices between 3rd and first vertex. In that case, X1+X2+X3 = 5 -------------- (1)
the number of ways vertices can be chosen will be equal to the number of solution of the eq1 sbove. where x1,x2,x3 > 0
Let x1 = y1 + 1, X2 = Y2 + 1, X3 = Y3 + 1.
then eq 1 changes to Y1+Y2+Y3 = 2, (y1,Y2,y3>=0)
And the number of ways of solving eq2 (by the book) is (2+3-1)C(3-1) = 4C2 = 6.
Also the number of vertices can be chosen in 8 ways ...
so total num of ways = 6*8 = 48.
But the answer given in the book is 16 & their approach is also different.

Any feedback on where my approach is wrong ?
If you want i can also post the solution given in the book but please also hav a look at my approach above and please tell me where i am mistaken. Thank you.

[yogendra]

Originally Posted by Durba

This is a common Q, however i am stuck at the approach. Why 1 method works and the other doesnt.
Q > Find the number of triangles whose vertices are vertices of an octagon but none of the sides of such triangle are taken from the sides of the octagon.

My approach >
the Q can be visualized as the number of ways of choosing vertices so that no two are consecutive. If the vertices are a,b,c,d,e,f,g,h and i pick "a" then let there be X1 vertices between "a" and the next vertices. Similarly X2 be the number of vertices b/w 2nd & 3rd vertices and X3 be number of vertices between 3rd and first vertex. In that case, X1+X2+X3 = 5 -------------- (1)
the number of ways vertices can be chosen will be equal to the number of solution of the eq1 sbove. where x1,x2,x3 > 0
Let x1 = y1 + 1, X2 = Y2 + 1, X3 = Y3 + 1.
then eq 1 changes to Y1+Y2+Y3 = 2, (y1,Y2,y3>=0)
And the number of ways of solving eq2 (by the book) is (2+3-1)C(3-1) = 4C2 = 6.
Also the number of vertices can be chosen in 8 ways ...
so total num of ways = 6*8 = 48.
But the answer given in the book is 16 & their approach is also different.

Any feedback on where my approach is wrong ?
If you want i can also post the solution given in the book but please also hav a look at my approach above and please tell me where i am mistaken. Thank you.

plz post solution from book too...
only flaw i can find is num of ways will be 6 not 6*8