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permut&combi
by karan batra - Saturday, 21 July 2012, 06:02 PM
 
Can somebody provide detailed solution to this problem , thanx in advance
Q) In how many ways can the letters of the word "SUBJECT" be placed in the squares of the figure shown below so that no row remains empty?

Plz give detailed solution wink
Re: permut&combi
by deepak kushwaha - Saturday, 21 July 2012, 06:43 PM
  each row should have atleast one alphabet ...

case 1 . top row 1
second row 1
third row 5
soln
7C2*2*2 * 6C5*5!

case 2 top row and second row 3 alphabet
bottom row 4 alphabet

soln
7C3 * 4C3 *3! * 6C4 * 4!


case 3
top two row 4 alphabet
bottom row 3 alphabet

7C4 * 4! * 6C3 * 3!

ans
= case1+ case2+case3
Re: permut&combi
by Mohit Sharma - Saturday, 21 July 2012, 06:44 PM
  Hi karna,
My take will be,
It says no row remains empty.
In first row= C(2,1)*C(7,1)
In second row= C(6,1)
In third row we would chose the four single boxes= C(5,4)*!4
Last character=C(4,1)
Ans= 2*7*6*120*4=40320
Please tell me if i m wrng.
Thanks,
Re: permut&combi
by rakshit surana - Saturday, 21 July 2012, 08:40 PM
  cases will be (1,1,5) , (1,2,4) , (2,1,4) , (2,2,3)
case I
C(7,2)*C(2,1)* 2*2*C(6,5)* 5!
C(7,2) we select two balls for first two balls
C(2,1) for choosing one row from first two..so one ball goes in one and other automatically in second.
2*2 for because in the first two rows , each ball has two squares to go
C(6,5) for choosing any five out of six squares from third row and arranging in 5! ways

case II
C(7,1)* 2*C(6,2)* 2! * C(6,4) * 4!

case III
same as case II

case IV
C(7,2)* 2! * C(5,2) * 2! * C(6,3)* 3!

sum up all the cases..
Re: permut&combi
by rakshit surana - Saturday, 21 July 2012, 08:46 PM
  sir, your quant lessons again puzzled my mind

Q no of solutions of xyz = 120
i dint got your concept
Re: permut&combi
by karan batra - Saturday, 21 July 2012, 09:32 PM
  unfortunately mohit and everyone all of ur answers are wrong , the correct answer is given as 524160.

I hope KAMAL SIR will provide us with detailed solution.
Re: permut&combi
by 50 cent - Sunday, 22 July 2012, 01:05 PM
  Yes the ans. given is correct. There would be 4 cases for this sum and when you subtract the not possible ways from the total ways you will get the correct ans.  and if you are trying with traditional ways then also the ans would be the same. Dont try and do it sticking to the formula. Be a bit logical and you will definitely crack the combinationsmile
Re: permut&combi
by karan batra - Sunday, 22 July 2012, 01:49 PM
  Hey Siddharth if  u are so sure , why not u tell the answer , i have tried all cases and the nearest answer i am getting by subtracting all the possible cases with total no. of ways is 534205 and the correct answer is 524160. If u get this answer then kindly tell us ur method/approach.
Re: permut&combi
by 50 cent - Monday, 23 July 2012, 10:21 AM
  Hi all,smile

There will be 4 cases in total:

CASE-1: When there are 2 letters in the row 1& 2 and remaining in row 3
so solving this will give  (2C2*7*6) *(2C2*5*4) * (6C3*3!) = 100800

CASE-2: When there are 2 letters in the row 1& 1 letter in row 2 and remaining in row 3
so solving this will give (2C2*7*6) * (2C1*5) * (6C4*4!)  = 151200

CASE-3: When there 1 letter in the row 1& 2 letters in row 2 and remaining in row 3
so solving this will give (2C1*7) * (2C2*6*5) * (6C4*4!) = 151200

CASE-4: When there 1 letter in the row 1& 1 letter in row 2 and remaining in row 3
so solving this will give 2 {(2C2*7) * (2C2*6)} * (6C5*5!) = 120960.
 
So the total ways would be 100800+151200+151200+120960 = 524160
Re: permut&combi
by rakshit surana - Monday, 23 July 2012, 12:47 PM
  mene upar jo solution post kia he usse sahi hi toh ans aa rha tha,,fir q bola galat aa rha he ans, none of them had solved sad
Re: permut&combi
by sahil sharma - Thursday, 26 February 2015, 08:47 AM
  Total answer is 7!*104
Case 1: Where we place one letter in row one, one in row two and 5 in row 3. total ways = 2*2*6C5
Case 2: One letter in row and 2 in row 2 and 4 in row 3
total ways = 1*2*6C4
Case 3 : 2 letters in row and 1 in row 2 and 4 in row 3
total ways = 1*2*6C4
Case 4 2 letters in row 1 and 2 and 5 in row 3
total ways = 1*1*6C3

IN all these cases we will have 7! ways to arrange the letters

Hence total ways = (case1+case2+case3+case4)*7!

= 104*7!
= 13*8!