Latest TCS Aptitude Question SOLUTION: 18 guests have to be seated half on each side of a long table.four particular guests desire to sit on one particular and three others on other side.the no. of ways in which sitting

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18 guests have to be seated half on each side of a long table.four particular guests desire to sit on one particular and three others on other side.the no. of ways in which sitting arrangement can be made?

Asked In TCS charan (11 years ago)

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Dinesh Gundapaneni  (11 years ago)

four on one side and three on other side.. hence guests number is limited to 11 as 7 persons are already fixed, since 18-(4+3)=11.
--->now consider the side with four people, hence we have to fill five more places,
selection= 11C5 arrangement= 9! (since the guests are fixed to only sides not the places on that particular side)
----> consider the other side we already have 3 people fixed and also remaining are 6 guests as already five are selected before for other side.
hence only arrangement= 9!

total= 11C5*9!*9!

Is this solution Helpfull? Yes (19) | No (4)  

aditya mangalampalli  (11 years ago)

(2c1)*(9p4)*(9p3)*(11p1)

=============>
2c1--->select a side
9p4------>4people can be arranged in 9 seats(one side) in 9p4 ways
9p3------>3people can be arranged in 9 seats(other side) in 9p3 ways
11p11----->remaining 11 people can sit any where so 11 people in 11 seats therefore 11p11 or 11!

Is this solution Helpfull? Yes (9) | No (2)  

amrendra kumar  (11 years ago)

the 4 person on one side can be seated in 9C4*4! ways.
the 3 person on the other side can be seated in 9C3 *3! ways.
now the two sides has to be multiplied by 2!
the remaining 11 person can be seated in 11! ways...

so total = 9C4*4!*9C3*3!*2*11!....

Is this solution Helpfull? Yes (4) | No (4)  

Sridhar Reddy  (11 years ago)

11c5*9!*9!*2 is the answer

Is this solution Helpfull? Yes (3) | No   

Akshay jain  (11 years ago)

3 persons will sit on 1 side and 1 person on other side.
now we have to select 6 persons for 1st side=14C6(which is equal to 14C8 for other side)
now on both sides 9 persons can be arranged in=9! ways
so total no. of ways= 14C6*9!*9!

Is this solution Helpfull? Yes (1) | No (3)  

mohd zubair khan  (11 years ago)

IF IT IS A RECTENGLAR TABLE THEN IT CAN BE SOLVED AS FOLLOWS
FIRST 9 PERSONS WILL SIT ON EACH SIDE OF TABLE
THEN SELECT OF 9 PERSONS FROM 18 PERSONS = 18C9 WAYS
NOW EACH 9 CAN BE SEATED AS= 9! X 9!
AFTER REST 4 GUESTS CAN BE SATED AS BETWEEN THE SPACE CREATED AMONG THESE PERSON
OUT OF IT ONE PERSON WILLING TO SIT ON A PARTICULAR SPACE SO REMAINING SPACES=21
SO THIS CAN BE DONE = 18C9 x 9! X 9! x 21C3 (ANS)

Is this solution Helpfull? Yes | No (19)  

vikash kumar  (11 years ago)

if it is a rectangular table then each side 9 guests can sit.
on one side of table always 4 particular guest will sit together so they can be sit in=6!*4P1 WAY.
ON the otherside always 3 will always sit together so the can be sit in=7!*3P1 WAY
SO ALLTOGETHER THEY CAN SIT IN=(6!*4P1)*(7!*3P1)WAY=43545600WAY

Is this solution Helpfull? Yes | No (5)  

Honey  (11 years ago)

4!*3!*11P5*11P6

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abhishek mishra  (11 years ago)

9p4*9p3*11

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