Latest TCS Aptitude Question SOLUTION: The crew of a rowing team of 8 members is to be chosen from 12 men (M1, M2, …., M12) and 8 women (W1, W2,…., W8), such that there are two rows, each row occupying one the 2 sid

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The crew of a rowing team of 8 members is to be chosen from 12 men (M1, M2, …., M12) and 8 women (W1, W2,…., W8), such that there are two rows, each row occupying one the 2 sides of the boat and that each side must have 4 members including at least one women. Further it is also known W1 and M7 must be selected for one of its sides while M2, M3 and M10 must be selected for other side. What is the number of ways in which rowing team can be arranged.

Asked In TCS Ramya (10 years ago)

Unsolved Read Solution (17)

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Read Solution (17) :

kumar ishan  (10 years ago)

Side One: Total Number 4. M2, M3 and M10 are must. This leaves one seat for woman. Out of the eight women, W1 should be on side two. That leaves remaining 7 women from whom one woman can be selected in 7 ways.

Side Two: Total Number 4. W1 and M7 are must. That leaves two remaining seats that can be filled exclusively by either women or one woman and one man. We have 6 remaining women and the two seats can be filled in 6C2 ways ie 15 ways. Suppose we fill one man and one woman this can be done in the following manner. Since there are 6 women and 8 men available the filling can be done in (6C1)*(8C1) = 48 ways.
and only 2 mens selected then 8C2=28

Thus the remaining seats in the boat can be filled in 7 + 15 + 48+28 = 98 ways.

Is this solution Helpfull? Yes (27) | No (4)  

lucky  (10 years ago)

7C1*14C2*4!*4!
second side will be definitely contain woman as there should be one women on each side so no of ways 7C1
and remaing 2 places we have to fill from 14 remaining people ways 14C2 and each side can be arranged in 4! ways

Is this solution Helpfull? Yes (20) | No (16)  

sonali soni  (10 years ago)

7C1*14C2*4!*4! NOW, since there are two rowa and rows can be arranged in 2 ways. therefore..(7C1*14C2*4!*4)*2

Is this solution Helpfull? Yes (20) | No (5)  

Shakti Panda  (9 years ago)

We need two person for one side and 1 women for the another side. We select that women in 7 ways. Now that second side people can sit in 7x4! ways.
Now for the first side we need two people from the remaining 14. So this can be done in 14C2 ways and this side people can sit in 4C2×4! ways.
Again the first group may take any of the two sides. So total ways are 2×7×4!×14C2×4!

Is this solution Helpfull? Yes (9) | No   

Nivetha  (9 years ago)

7C1*14C2=637 ways.

Is this solution Helpfull? Yes (7) | No (5)  

Saatvika Dhumeja  (10 years ago)

Side One: Total Number 4. M2, M3 and M10 are must. This leaves one seat for woman. Out of the eight women, W1 should be on side two. That leaves remaining 7 women from whom one woman can be selected in 7 ways.

Side Two: Total Number 4. W1 and M7 are must. That leaves two remaining seats that can be filled exclusively by either women or one woman and one man. We have 6 remaining women and the two seats can be filled in 6C2 ways ie 15 ways. Suppose we fill one man and one woman this can be done in the following manner. Since there are 6 women and 8 men available the filling can be done in (6C1)*(8C1) = 48 ways.

Thus the remaining seats in the boat can be filled in 7 + 15 + 48 = 70 ways.

Is this solution Helpfull? Yes (5) | No (7)  

keerthana  (7 years ago)

row 1: W1 M7 __ __
row 2: M2 M3 M10 __
In row 2 : 7C1=7... so no of possibility is 7*4!
in row 1: 2 remaining spaces may take i) 2 men: 8C2=28 or ii) 2 women 6C2 or iii) 1 man and 1 women : 8C1*6C1=48
so.... (28+15+48)*4!
At last... 91*7*4!*4! ways
but we can arrange the members in two rows in two ways... so ans is 2*91*7*4!*4!

Is this solution Helpfull? Yes (4) | No   

Syed Mohd Fareed  (8 years ago)

We need two person for one side and 1 women for the another side. We select that women in 7 ways. Now that second side people can sit in 7x4! ways.
Now for the first side we need two people from the remaining 14. So this can be done in 14C2 ways and this side people can sit in 4C2×4! ways.
Again the first group may take any of the two sides. So total ways are 2×7×4!×14C2×4!

Is this solution Helpfull? Yes (3) | No   

prathyusha  (8 years ago)

Side one: 7C1
Side two: 6C2 + (6C1*8C1) + 8C2
No. of ways = 7+15+48+28 =98

Is this solution Helpfull? Yes (2) | No   

pre  (10 years ago)

In first side,
11 men and 7 women are left and the condition of atleast one women is fulfilled by the given condition thus men can be selected in 11C2 ways similarly in second case there is no women who already selected so it can be selected in 7C1 ways the total answer is 11c2*7c1.

Is this solution Helpfull? Yes (1) | No (2)  

vishal kumar singh  (10 years ago)

We need two person for one side and 1 women for the another side. We select that women in 7 ways. Now that second side people can sit in 7x4! ways. Now for the first side we need two people from the remaining 14. So this can be done in ${}^{14}{C_2}$ ways and this side people can sit in ${}^{14}{C_2} times 14!$ ways. Again the first group may take any of the two sides. So total ways are $2 times 7 times 4! times {}^{14}{C_2} times 14!$

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

Rani  (8 years ago)

We need two person for one side and 1 women for the another side. We select that women in 7 ways. Now that
second side people can sit in 7x4! ways.
Now for the first side we need two people from the remaining 14. So this can be done in 14C2 ways and this side
people can sit in 4C2 × 4! ways.
Again the first group may take any of the two sides. So total ways are 2 × 7 × 4! × 14C2 × 4!

Is this solution Helpfull? Yes (1) | No   

Vibhuti Sharma  (8 years ago)

We need two person for one side and 1 women for the another side. We select that women in 7 ways. Now that second side people can sit in 7x4! ways.
Now for the first side we need two people from the remaining 14. So this can be done in 14C2 ways and this side people can sit in 4C2×4! ways.
Again the first group may take any of the two sides. So total ways are 2×7×4!×14C2×4!

Is this solution Helpfull? Yes (1) | No   

Deepak Chempazhi  (10 years ago)

428
2!(12C2 +12C1*8C1+8C2)+3!(8C1)

Is this solution Helpfull? Yes | No (18)  

Chandan Sharma  (10 years ago)

(7C2*8C2)+7C1

Is this solution Helpfull? Yes | No (10)  

Raju Sharma  (10 years ago)

11C2 *4!+8C1*4!

Is this solution Helpfull? Yes | No (3)  

DEVENDRA VIJAYRAO BORASE  (7 years ago)

We need two person for one side and 1 women for the another side. We select that women in 7 ways. Now that second side people can sit in 7x4! ways.
Now for the first side we need two people from the remaining 14. So this can be done in 14C2 ways and this side people can sit in 4C2×4! ways.
Again the first group may take any of the two sides. So total ways are 2×7×4!×14C2×4!

Is this solution Helpfull? Yes | No   

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