11@ rishan.... the question says that there should be no boys between two girls...
but in ur answer u have put two boys in between 2 girls...
u have gone wrong there...
@ashish..... see if we have arrange boys and girls in the manner, the question reads...
we have to put all the girls together.... otherwise this cant be done
so now lets arrange the boys...
_B_B_B_B_B_B_
SEE there are 7 places in which the girls can sit...
so, it can be selected in 7C1 ways..
as, the boys can interchange positions between themselves... (6!)
and the girls can do the same among themselves...(4!)
thus the number of ways in which this can be done = 6!* 7C1*4!...
ISNT IT???
11
Joydoot ghatak
·2010-10-24 10:58:10
i did the same thing nishant sir...
621.In how many ways can 6 boys and 4 girls sit in a row so that no boy is between two girls?
Lets make a block of the 4 girls.. they have to be seated together so that no one comes in between..
So we have 7 items.. (6 boys and a block of 4 girls)
so the no of ways to arrange them is 7!
over that the girls themselves can arrange in 4! ways..
So the answer to the first one will be 7!.4!
11
Joydoot ghatak
·2010-10-22 20:42:32
are u sure??? pls... check it once more...
because me and rishan getting the same answer...
1ya joydoot u r right!!
that means in my solution it will only be the first case!!!
11
Joydoot ghatak
·2010-10-15 02:46:34
2)
if the number is to be divisible by 4.
the last digit should be divisible by 4.
this can be done only if the last two digits be...
12,
24,
32,
44 and
52.
other multiples of 4 are eliminated because their digit do not lie between 1 to 5
therefore last 2 digits can be filled in 5 ways..
the first place can be filled in 5 ways...
again, the second place can be filled in 5 ways...
similarly the third place...
(i.e. from 1 to 5.... as digits can be repeated...
so the answer should be = 5*5*5*5 = 625.
is the answer correct jangra???
1
rishan chattaraj
·2010-10-20 01:23:18
1)two cases
a) when both the girls and the boys are separate
then 7!*4!
b) when girls and boys mix up
then
two cases
1)GBBGBBGBBG then 6!*4!
2)BGBBGGBBGB THEN ALSO 6!*4!
FINALLY ANS IS SUMMATION OF THESE
30@Joydoot - Can you explain your answer to the 1st question?
3yup...joydoot!! correct u r!! seems that u have confronted such problems earlier too!! :)
11
Joydoot ghatak
·2010-10-15 03:21:05
3)
there should be 5*4*3*2*1 =5! = 120 such numbers...
each place has 120/5 =24 1's , 24 3's ,24 5's , 24 7's and 24 9's...
thus the sum of the numbers in each of the places ( the unit's place, the ten's place, the hundred's place , the thousand's place and the ten thousand's place.... all have the same sum i.e.) = 24*(1+3+5+7+9) =600 .
therefore the total sum = 600 + 600*10 (for ten's place) + 600*100 (for hundred's place) + 600*1000 + 600*10000 = 6666600.
11
Joydoot ghatak
·2010-10-15 03:08:11
4)
THE NUMBER OF DIFFERENT SIGNALS THAT COULD BE SENT = 4n - 1
This is given as 1023 i.e. 45 - 1.
thus n = 5.
so 5 should be the answer.
11
Joydoot ghatak
·2010-10-15 03:01:58
1)
is the first answer 6!*7C1*4!
