| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 25 Jun 2007 09:28:55 IST
|
|
|
Q1) EVERYBODY IN A A ROOM SHAKES HANDS WITH EVERY BODY ELSE.THE TOTAL NUMBER OF HANDSHAKES IS 66.THE TOTAL NUMBER OF PERSONS IN THE ROOM IS?
a)11 b)12 c)13 d)14
Q2)THE MAXIMUM NUMBER OF POINTS OF INTERSECTION OF 8 CIRCLES IS
a)16 b)24 c)28 d)56
Q3)ELEVEN ANIMALS OF A CIRCUS HAVE TO BE PLACED IN ELEVEN CAGES ONE IN EACH CAGE.IF 4 OF THE CAGES ARE TOO SMALL FOR 6 OF THE ANIMALS ,THEN THE NUMBER OF WAYS OF CAGING THE ANIMALS ARE:
a)304800 b)504800 c)604800 d)NONE OF THESE
Q4)THE NUMBER OF WAYS ONE CAN PUT 3 BALLS NUMBERED 1,2,3 IN THREE BOXES LABELLED A,B,C SUCH THAT AT MOST ON ONE BOX IS EMPTY IS EQUAL TO
a)6 b)24 c)42 d)18
Q5)THE RESULT(WIN,DRAW OR LOSS)OF 10 FOOTBALL MATCHES ARE TO BE PREDICTED.THE NUMBER OF DIFFERENT FORECASTS THAT CAN CONTAIN EXACTLY 5 CORRECT RESULTS IS:
a)2^5 b)10C5 c)10C5 * 2^5 D)NONE OF THESE
|
|
|
|
25 Jun 2007 09:37:55 IST
|
|
|
Hey Buddy I am trying ...1st and 4th are easy ..just need to calculate
The other ones are really tricky . I need some time. Why don't you ask these to the experts here ....they may answer better.
|
Ken
From: UNITED STATES, Green Bay, Wisconsin
 |
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|
|
|
Q1: total no: of shake hands = nC2 = 66 n (n-1) /2 = 66 ie n(n-1) = 66*2 = 11*12 no:of ppl = 12
|
KINDLY RATE ME 4 MY EFFORTS PLZZZZZZZZZ......... |
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
25 Jun 2007 09:45:34 IST
|
|
|
ans1. the ans is
nC2 =66
n(n-1)=132
or n =12
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
25 Jun 2007 09:50:50 IST
|
|
|
ans 2. 2 circles can intersect at atmost 2 points .... max no of pts can be obtained if no 3 circles intersect at one point...
no. of all possible pairs of circles 8C2
max
no. of points of intersection 2. 8c2=56
|
this reply: 7 points
(with 1
in 2 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
25 Jun 2007 09:51:05 IST
|
|
|
Q3: out of 11, 6 animals r big & 5 r small 4 cages r small ,7 r big arrange any 4 from 5 animals in 4 small cages & remaining animals in big cages no: of ways = 5C4 * 7C7 = 604800 opt(c)
|
KINDLY RATE ME 4 MY EFFORTS PLZZZZZZZZZ......... |
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
25 Jun 2007 09:53:53 IST
|
|
|
ans 3 ..
5 animals can fit into 4 cages... so no of ways is 5c4 . 4! .7!=604800
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
25 Jun 2007 10:00:10 IST
|
|
|
ans. 4 atmost 1 box is empty means either no enpty or 1 empty
no empty box no of wayz.. 3!=6
1 box empty... 3C1.2c1 3c2 1c1=18...
3c1 to select empty box... 2c1 to select box wid 2 balls... 3c2 to select the 2 balln 1c1 to put remainin ball in remaini box
total 24
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
25 Jun 2007 10:01:47 IST
|
|
|
Q4: balls can b arranged 1 in each box or 2 in 1 ,remaining in 1 , 1 left empty ways 2 arrange 1 ball each in a box = 3P3 = 6 2 keep 1 empty,first select 1 which is 2 b empty = 3C1 take any 2 balls from 3 = 3C2 choose a box from non empty boxes = 2C1 put 2 balls in selected 1 & remaining in other total ways = 3C1 * 3C2 * 2C1 = 18 total = 6 + 18 = 24
|
KINDLY RATE ME 4 MY EFFORTS PLZZZZZZZZZ......... |
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
25 Jun 2007 10:03:35 IST
|
|
|
these ques are very tricky literally im tryin to solve dem will ans soon but better option is trust the experts
|
" I've shed many tears. But after, I realize, why am I crying? No matter how bad it got, it could be worse. No matter how much you think it's not going to be ok, eventually it will be. You just have to let it be."
EXPECT MORE FROM URSELF THAN FRM OTHERS AS EXPECTATION FRM OTHERS HURTS U WHILE FROM UR SELF INSPIRES U A lot |
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
25 Jun 2007 10:03:50 IST
|
|
|
ans 5..
3rd option is right...
select 5 games whose results r wrongly predicted 10c5
then 4 wrong prediction u hv 2 options(as one is correct result out of three)
so total wayz 10c5. 2^5
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
25 Jun 2007 10:09:55 IST
|
|
|
each prediction can b either correct or wrong a prediction can b correct only in 1 way,wrong in 2 ways 2 make 5 correct & 5 wrongs select 5 correct out of 10 predictions = 10C5 no:of ways = 10C5 * 2^5(wrong can b in 2 ways)
|
KINDLY RATE ME 4 MY EFFORTS PLZZZZZZZZZ......... |
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
14 Sep 2007 08:55:26 IST
|
|
|
I am not a member of GoIIT... This is just a fake ID I had to make, so that I could reply to the third question... Since 4 cages are too small for 6 of the 11 animals, these 6 have to be put in the remaining 7 cages, and that can be done in : c(7,6) X 6! ways = 7 X 6! ways = 7! ways. (putting 6 in 7) (arranging 6) While the remaining five animals can be put in the remaining 5 cages in 5! ways. So, total number of ways = 7! X 5! ways = 604800 ways. Hope you got it. (That's a fake one too, lol) I would be glad to help you out.
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
14 Sep 2007 13:25:32 IST
|
|
|
answer to the third question
five smaller animals can fit in the four cages
so number of ways of choosing = 5C4
now number of permutations = 5C4*4!
now the remaining seven cages have to be filld with seven animals
so number of permutations = 7C7*7!
so total = 5C4*4!*7C7*7! = 604800
|
I like to be myself. |
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
14 Sep 2007 13:31:41 IST
|
|
|
Moderation Team
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
