IIT JEE , AIEEE Forum - Mathematics , Algebra

coolkrazy007

1)the no. of square on a chess board is-
options:a)64,b)160,c)224,d)204

2)the total no. of seven digit nos. the sum of whose digits is even is-
options:a)9000000,b)4500000,c)8100000,d)none of these

3)the no. of different word groups formed by taking at least one letter from the letters of the words 'RANDOM','NOBLE,and 'INTEGRAL' are-
a)248031,b)255,c)230975,d)498015

4)20 persons were invited for a party.in how many ways can they nd the host be seated at a circin an ular table?in how many of these ways will two particular persons be seated on either side of the host?

5)find the number of ways in wch 10 candidates A1,A2,A3,.....A10 can be ranked a)if A1 and A2 are next to each other,and
b)A1 is just above A2
C)if A1is always above A2

6)in an examination,the maximum marks for each of the three papers are 50 each.maximum marks for the fourth paper are 100.find the no. of ways in wch the candidate can score 60% marks in aggregate.

answers:
1)D
2)B
3)D
4)20!,2(18)!
5)a)2.9!, b)9!, c)10!/2
6)110551

explain wid pics........

akhil_o

Already solved earlier

Ans to first question-204
method:

no of squares of dimension:
1x1 = (8*8)
2x2 = (7*7)
3x3= (6*6)
.
.
.
8x8= 1*1

so adding al these we get

12+22+...82
=8(9)17/6
=204

coolkrazy007

if no. of rectangles are there in Q1 instead og sqres den wat do i do>?

ramyani

1st prob

click here

akhil_o

For no. of rectangles

1x1- 8*8

1x2- 8*7

1x3- 8*6

.

..

1x8- 8*1

similarly for 2

2x 2- 7*7

2x3- 7*6

.

similarly for the rest tll 8

8x8 = 1*1

hence total no of rectangles

is

8(1+2+3+...8)

+7(1+2+3+...7)

+6(1+2+3+...6)

..

+1(1)

this gives

8*8*7/2+ 7*7*6/2+ 6*6*5/2...2*2*1/2

add the series and get the answer

rate if useful

nudge if doubtful

computer001

Re:permutation and combination

coolkrazy007

can u add me to ur yahoo messenger..

coolkrazy007@yahoo.com

i need to ask u alot...me still confused..plzzzzzzzz

Greatdreams

How can the number of rectangles on a chess board be 771??

I think the no of rectangles will be 9C2 * 9C2 = 1296

6th one is good

but tremendous calculations

total marks = 2*50+100 = 250

so 60% marks means 150

So we have to apply multinomial theorem here,

the different ways are :

is coefficient of x150 in (1+x+x²+x³+....+x⁵⁰)³
*(1+x+x²+x³+x⁴.....+x¹⁰⁰)

So we have (1-x⁵¹/1-x)³*(1-x¹⁰¹/1-x)

or (1 - x⁵¹)³*(1-x)¹⁰⁰*(1-x)^-4

so (1-3x⁵¹+3x¹⁰²-x¹⁵³)*(1-x¹⁰¹)*(1+4x+4*5/2x²+.....)

solve this and you will get it...

computer001

2nd prob is simple
for sum of digits to b even u need to worry only abt last digit
the 1st 6 digits can b formed in 9*10*10*10*10*10 ways i.e 1st digit other than 0..next 5 digits nething fron 0 to 9...now for ne given combination the sum of 1st 6 digits may b even or odd..suppose it is odd u can select the last digit in 5 ways to make it even(1 or 3 or 5 or 7 or 9)which is 5 ways
similarly if the sum of 1st 6 digits is even u can find 5 poss for the last digit..
among the 1st 6 digits num of even will be 9*10*10*10*10*10/2 and same for odd also.. and for each of them u multiply by 5 for last digit..
so total ways =5*9*10*10*10*10*10/2 + 5*9*10*10*10*10*10/2
=4500000

computer001

my ans also tallies wid akhil or whoever..

computer001

no sorry i made a mistake...
ans is (1+2+3+4+5+6+7+8)(1+2+3+4+5+6+7+8)
which is 1296...
@ greatdreams:
made a mistake as i thot il have overlap..

computer001

@ greatdreams:
n do u agree wid my 2nd sum soln

akhil_o

4)
a) grouping A1 and A2 as one
we have 9! ways. now A1 and A2 can be exchanged
hence 9!*2

b)
again group A1 and A2
hence 9! ways.this time positions of A1 and A2 r fixed

c)
total ways is 10!.Now in half these cases A1 >A2 and in the rest half vice versa
hence we take only the second half
10!/2

akhil_o

already done 2 of these in ur other post man! where r u?

Anyway no of rectangles comes out to

(n)(n+1)(3n²+7n+2)/24

where n=8

substituting we get no of rectangles =750

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