The number of ways in which three letters be posted in four letter boxes in a village, if all the three letters are not posted in the same letter box.
A) 64
B) 60
C) 81
D) 78
E) None of these
GMAT Question of the Day: Permutation and Combination
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If letters are posted induvidually…
Four boxes can be filled with first letter, second letter, third letter and blank can be filled in 4X3X2X1 = 24 ways of posting
If two letters are posted in one box and one in another box, two will be blank.So (combo of two, third letter, blank and blank can be done in (4X3X2X1)/2X1=12; because blanks are duplicated here the formula here is 4C2. The combo of two can be selected in 3 different ways; so the total number of ways of posting with two letters in one box is 3X12 = 36.
All letters cannot be posted per the question.
So the total number of ways 3 letters can be posted in four boxes is 24+36 = 60.
Choice is B
Dear Krishna Mohan,
That is not the correct answer.
Can you please post the answer and explain. Thanks.
Dear Krishna Mohan,
Sorry for mistake. we appolisise for your inconvenience.
Oh boy! i was struggling where i went wrong and went around searching. Thanks for letting me know.
so is 60 the correct answer?
Dear Krishna,
The answer is correct, but I guess the logic osf getting 36 is a little different:
There are following possibilities of dropping letters in mail boxes:
A1 B2 ; A1 C2; B1 A2; B1 C2; C1 A2; C1 B2 which means choosing 2 boxes out of 4 (multiplied by 6 possible combinations) i.e. (4C2 * 6) = 36, making the answer as 24+36 = 60
b
There r 3 letters and 4 boxes. If they dropped one letter in one box 4p3 ways can be possible .i.e (4!/(4-3)!= 24) and
secondly 2 letters dropped at a time the we can drop those letters in the 3x ( 4p2) = 36
so total ways of posting the letters r 60
hey, shanti, can u explain, how did u get the ans .
I solved this using logic:
Each Letters can take 4 boxes : 4x4x4=64 ways
All 3 cannot be in the same letter box:So eliminate 4 ways!
64-4=60!
Thanks for the answer shanti.
b
San, I solved the answer using the same method as you did…
Solution Usinf fundamental Law of counting
ANS-60
Acc to FLC,
no of ways 1st letter can be dropped=4
no of ways 2nd letter can be dropped=4
no of ways 3rd letter can be dropped=4
smulateously all three can be dropped =4*4*4=64
but in this 64, there are 4 possible ways of dropping all letters in each box, where all three can be dropped and no letter in any other box.
So ANS=64-4=60
No of ways in which 3 letters can be posted into 4 boxes is 4^3
No of ways in which 3 letters cant be put in the same box is 4(since 4 boxes)
so no of ways in which 3 letters can be posted into 4 boxes any by not posting all the 3 letters in the same box is 4^3 -4 = 60 (B)
4*4*4-4
choice is B
B
friends,60 is the right answer.
Total no of ways=4^3=64
No of ways to put all three in one letterbox=4
So,answer is 64-4=60
81 is the answer.if not then mail me the correct ans
since there are four postbox’s and three letters ..then for the first letter we have 4 options for second 3 and for the third we have 2..that means 4*3*2=24 ways..
now we can have two sets one of two letters and one of single letter..
so picking two letters of three is 3C2 =3
and for this set of two letter we have 4 letterbox..and for the single letter we have 3 letterboxes…so 3*4*3=36 ways we can post the one set of two letters and a single letter..
therefor the total combination is 36+24=60ans
thanks sam, you are genius, please keep me posted on more simplied logic for combinatorics
can any one explain in detail pl.
B
81 is the ans
60 is the ans
The right answer is 60.
4*4*4-4
The answer is 60 ways/
B
B..
E. I think 12 is the right answer
B…
in case there are still doubts….
i started calculating the number of ways i can deposit 3 letter in 4 mailboxes. afterwards I substracted the number of alternatives were the 3 letters were posted in the same mailbox .
step 1.-(number of mailboxes)^(number of letter)=4^3=64
step 2.-number of alternatives were the 3 letters were posted in the same mailbox=4 (3 letters in 1, 3 letters in 2, 3 letters in 3, 3 letters in 4)
64-4=60
the correct answer is “B”.
2 cases
1st Case: 3lletters in 3different boxes so 4C3x3!=240r we can say we choose 3 boxes from 4 letter boxes and then arrangement of 3 lletters can be made in (A,B,C)3!ways ie posted in 3 boxes
2nd Case:2letters in 1box and 1letter in another box.Again choose 2 letter boxes from 4boxes ie 4C2 and grouping of 3letters in 2 groups can be made in (2letters and 1 letter)3C2 ways now these 2 groups can be arranged among themselves in 2! ways.
therefore 4C2x3C2x2!=4x3x3=36
Total=36+24=60
Ans:B
Gotta be E, you guys. Someone prove me wrong. No formulas needed, just common sense.
Letters by themselves first. Extrapolate from each example.
ABC_, ACB_, AB_C, AC_B, A_CB, A_CB (A in box #1) = 6 permutations per letter per box.
6 perm per box per letter already = 72.
Two letters per box = 12 permutations per letter per box.
It’s way over 80. E is correct.
B is correct.
Number of ways 3 letters can go in 4 boxes : 4x4x4 = 64
Number of ways all 3 letters can go in one box = 4
So number of ways,if all the three letters are not posted in the same letter box = 64 – 4 = 60