Latest Elitmus Aptitude Question SOLUTION: In a theatre all rows have same no. Of . One family buys 10 tickets of first two row. 3 members want to sit in first row and 2 in second row. In how many ways they can be seated.

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(#M40157810) ELITMUS Numerical Ability question Keep an EYE

In a theatre all rows have same no. Of . One family buys 10 tickets of first two row. 3 members want to sit in first row and 2 in second row. In how many ways they can be seated.

Asked In Elitmus Sniper (7 years ago)

Unsolved Read Solution (11)

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Read Solution (11) :

abdullah imran  (7 years ago)

If out of 10 tkts r ticts are from 1st row then 10-r tkts in 2nd row. So ----->rC3 * (10-r)C2

So required Answer----->
3C3 * 7C2 + 4C3 * 6C2 + 5C3 * 5C2 + 6C3 * 4C2 + 7C3 * 3C2 + 8C3 * 2C2

Is this solution Helpfull? Yes (30) | No (2)  

paras  (7 years ago)

10c3x10c2= 5400

Is this solution Helpfull? Yes (13) | No (5)  

Shubham jha  (7 years ago)

after arranging 3 persn ...in first row .nw we hv 7 sit left in which we can accomodate 2 pesn..inthis way we hv following ways..10c3+7c2

Is this solution Helpfull? Yes (6) | No (2)  

nilu   (7 years ago)

10c3 * 7c2 + 10c2 * 8c3

Is this solution Helpfull? Yes (4) | No (2)  

abdullah imran  (7 years ago)

If out of 10 tkts r ticts are from 1st row then 10-r tkts in 2nd row. So required ans----->rC3 * (10-r)C2

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

Reetika Maurya  (5 years ago)

Out of 10 people choose 3 for first row and arrange them , so number of ways = 10C3 * 3!
Out of remaining 7 people choose 2 people for 2nd row and arrange them, so number of ways = * 7C2 * 2!
Now remaining 5 people can sit any row and can be arranged in 5! ways

So total number of ways = 10C3 * 3! * 7C2 * 2! * 5!

Is this solution Helpfull? Yes (3) | No   

avee  (7 years ago)

let 3 people n 2 people sit in their rows so ans will be 5c2*5! + 3c3*5!

Is this solution Helpfull? Yes (2) | No   

Aseem  (7 years ago)

since in first row 3 permanent member make 6 possible arrangement and and 2 in the second row make 2 possible arrangement hence we left with 5 people who can sit anywhere so they make 5! =120 possible arrangements so answers will be 120+2+6= 128

Is this solution Helpfull? Yes (1) | No   

  (3 years ago)

10c3*3!*7c2*2!*5!+10c2*2!*8c3*3!*5!

Is this solution Helpfull? Yes (1) | No   

Anurag  (7 years ago)

3 members to sit in 1st row = 10C3
2 members to sit in 2nd row = 10C2
Now, remaining members = 5 can be seated in = 7C5 + 7C4*8C1 + 7C3*8C2 + 7C2*8C3 + 7C1*8C4 + 8C5
Therefore, finally number of ways = 10C3 * 10C2 * {7C5 + 7C4*8C1 + 7C3*8C2 + 7C2*8C3 + 7C1*8C4 + 8C5}

Is this solution Helpfull? Yes | No (2)  

Survesh Jha  (3 years ago)

Total number of people = Total number of tickets = 10
Each row has 5 seats (Deduced from the statement.)
3 wants to sit in first row and 2 wants to sit in second row . (So assume that you've told them to sit ).
10 = 5 + 3 + 2.
now,
select two members from remaining 5 members (5C2) and make them sit in first row and arrange them by 5!.
similarly select remaining members (3C3) and make them sit it second row and arrange them by 5!.

so net answer will be

(5C2 * 5!) * (3C3 * 5!).

note that fundamental principal of multipication will be used because if any of the member change it's position its a whole new arrangement.

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