@amresh_maverick said:1> How many 3's will be there in a book having 700 pages ?2> 5 couples in a party .how many ways 4 persons can be seated on a table of 4 chairs such that no two husband wife are on the table ?
1> 240
2>192?
@amresh_maverick said:1> 3x +y +z=40 where x ,y,z are +ve integers , no of sol ?2> no distinct rectangles having integral sides such that perimeter is an integer less than 100 cm
2(l+b)
l+b
l+b = 2 -> 1 solution
l+b = 3 -> 1 soution
l+b = 4 -> 2 solutions
l+b = 5 -> 2 solutions
.
.
.
l+b = 48 -> 24 solutions
l+b = 49 -> 24 solutions
total = 2(1+2+...24) = 600 solutions
@amresh_maverick said:all squares are rectanglesL=1 , B=48 and B=1 , L=48 but we have to find no of distinct rectangles
l=b=49
24
48-24
47-23
36-23
24
48-24
47-23
36-23
1+2+...+24 *2=600
@amresh_maverick
subtract squares = 1176 - 24 = 1152
remove repetition = 1152/2 =576
add the squares back = 576 + 24 = 600
@amresh_maverick said:
2> 5 couples in a party .how many ways 4 persons can be seated on a table of 4 chairs such that no two husband wife are on the table ?
case 1: 1 husband wife are on the table they can be selected in 5 ways.
the remaining 2 people can be selected in 8*6=48 ways so totally 240 ways
case 2: no couple are on the table then 4 persons can be chosen in 10*8*6*4=1920
Now they can be seated in 4! ways so totally 2160*4!?
the remaining 2 people can be selected in 8*6=48 ways so totally 240 ways
case 2: no couple are on the table then 4 persons can be chosen in 10*8*6*4=1920
Now they can be seated in 4! ways so totally 2160*4!?
@amresh_maverick said:2> 5 couples in a party .how many ways 4 persons can be seated on a table of 4 chairs such that no two husband wife are on the table ?
EDIT:
1920 *4!
1920 *4!
@techgeek2050 said:@amresh_mavericksubtract squares = 1176 - 24 = 1152remove repetition = 1152/2 =576add the squares back = 576 + 24 = 600
though i stuck midway using this approach ,it is really short and refined..
Thanks for clarification...
Thanks for clarification...
@rnishant231 said:though i stuck midway using this approach ,it is really short and refined..Thanks for clarification...
any time bro. 
@amresh_maverick said:2> 5 couples in a party .how many ways 4 persons can be seated on a table of 4 chairs such that no two husband wife are on the table ?
is question me no two husband wife are on the table ........... it means ki not even one couple is allowed or one couple is allowed...........
@amresh_maverick said:1> How many 3's will be there in a book having 700 pages ?2> 5 couples in a party .how many ways 4 persons can be seated on a table of 4 chairs such that no two husband wife are on the table ?
1) 3ab -> 100
a3b -> 70
ab3 -> 70
but 333 is counted thrice, hence 240 - 2 = 238 times??
EDIT: misunderstood the second question, reworking.
@Dexian said:is question me no two husband wife are on the table ........... it means ki not even one couple is allowed or one couple is allowed...........
if polygamy is allowed then no couple should be considered 
@iLoveTorres said:if polygamy is allowed then no couple should be considered
i strongly condemn this.............
how many zeroes are present at the end of 25!+26!+27!+28!+30!?????
@pakkapagal said:how many zeroes are present at the end of 25!+26!+27!+28!+30!?????
25!(1+26+26*27+26*27*28+26*27*28+30)
number of 5 in 25! is 6....
rest will give one more 5 ...
so 7
number of 5 in 25! is 6....
rest will give one more 5 ...
so 7
@pakkapagal said:how many zeroes are present at the end of 25!+26!+27!+28!+30!?????
25!(1+26+27*26+28*27*26+30*29*28*27*26) = 10^6(xxx05)
so 7 zeros..
@Dexian said:25!(1+26+26*27+26*27*28+26*27*28*29+26*27*28+30)number of 5 in 25! is 6....rest will not give any 5...so 6
if you see internal term , it will sum up to unit digit 5
@pakkapagal said:no bhai...answer 6 diya hai but m getting 7 :/
@Dexian said:25!(1+26+26*27+26*27*28+26*27*28*29+26*27*28+30)number of 5 in 25! is 6....rest will not give any 5...so 6
bhai 7 hi hoga...