Latest TCS Aptitude Question SOLUTION: 18 guests have to be seated,half on each side of a long table .4 particular guests desire to sit on one particular side and 3 others on the other side .Determine the number of ways

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18 guests have to be seated,half on each side of a long table .4 particular guests desire to sit on one particular side and 3 others on the other side .Determine the number of ways in which the seating arrangement can be made

Asked In TCS ishwarya (11 years ago)

Solved subahn shaik Read Solution (15)

Is this Puzzle helpful?   (23)   (7) Submit Your Solution Permutation and Combination

Read Solution (15) :

Mrinal Mahanta  (11 years ago)

9p4*9p3*11!*2
4 guests one side out of 9 seats =9p4
3 guests another side out of 9 seats =9p3
remain 11 guest can sit in 11 seats any where =11!
the 3 and 4 perticular guest can change their side =2

Is this solution Helpfull? Yes (39) | No (13)  

Bharat  (11 years ago)

9p4*9p3*11!

Is this solution Helpfull? Yes (28) | No (8)  

subahn shaik  (11 years ago)

9p4*9p3*11!

Is this solution Helpfull? Yes (11) | No (3)  

Krishap Shah  (11 years ago)

11c5*6c6*9!*9! (ans)

The reason is simple, for the first table 4 particular people sits together and 3 for the second. Therefore 11 people out of 18 are left. so for the first table they can be selected in 11c5 ways. Similarly for the second table the no of combination is 6c6. Now the people can be arranged among themselves in 9!*9! ways.
Therefore total no. of arrangements=11c5*6c6*9!*9!

Is this solution Helpfull? Yes (11) | No (2)  

Keerthi  (11 years ago)

6!*4!*7!*3!*2
out of 9, 4 sit together.so total number of units are 6.(9-4=5 and 4 together as 1 unit).so they can be arranged in 6!*4! ways...similarly on another side, if 3 sit together 7!*3! ways. and 2! is for if they select an other side.

Is this solution Helpfull? Yes (7) | No (2)  

ankita gupta  (11 years ago)

9p3*9p4*11!

Is this solution Helpfull? Yes (3) | No (2)  

Rajat Jain  (9 years ago)

11 C 5 * 9! * 9 !......................................D

side 1
______________

side 2


18 guests
4 chooses to sit on side 1
3 chooses to sit on side 2

now 11 left
now 14 guest left
we have to choose 5 guests for side 1 in 11 C 5 ways
rest 6 will seat on other side
and now each side can be arranged in 9! ways
hence total arrangement = 11 C 5 * 9! * 9!

Is this solution Helpfull? Yes (2) | No   

Ashish Singh  (11 years ago)

bro 11! is in which manner come pls explain

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

Shatkratu Raj  (11 years ago)

9p4*9p3*11!*2

Is this solution Helpfull? Yes (1) | No (4)  

sneha sarkar  (11 years ago)

9p4*9p3*11!

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

Saurabh Anand Srivastava  (11 years ago)

!6*!4*!8*!3

Is this solution Helpfull? Yes | No (9)  

arunabha  (11 years ago)

9c7*9c6*4!*3!
9 on each side.
4 persons fixed so tak dat to b 1.
der fore no of ways is 9c6 ways * 4!
same on the odr side 9c7 * 3!

Is this solution Helpfull? Yes | No (11)  

amit ranakaant saha  (11 years ago)

at first take 4 guests in one side and likewise take 3 guests for other side.now 11 guests remains.so for first side where only 9 guests are allowed the remaining 5 guests will be taken out of 11 in C(11,5) ways , similarly for 2nd side where also 9 guests are allowed the remaining 6 guests will be taken out of 11 in C(11,6) ways.
in every side the guests can be arranged in 9! ways .
hence the solution is C(11,5)*C(11,6)*9!*9! ways=2.810*10^16

Is this solution Helpfull? Yes | No (2)  

NAMRATA KASHYAP  (9 years ago)

9p4*9p3*11!*2

Is this solution Helpfull? Yes | No (3)  

Anindya Mallick  (8 years ago)

11c5*9!*9!+11c6*9!*9!
=11c5*9!*9!+11c5*9!*9!
=2*11c5*9!*9!..

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