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Combination Problem

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Combination Problem

by agni_h1 » Wed Nov 03, 2010 10:09 am
Q.In how many ways can two identical lions can be put in eight identical cages such that the two lions are not in adjacent cages ?

No answer choices were given. I got the answer as 21 by taking one lion, one box and another lion as one unit. And then went on increasing the number of boxes between them.
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by Bharat » Wed Nov 03, 2010 4:23 pm
Answer: 21
a. total # of ways to put 2 identical lions in 8 identical cages= 8*7/2 = 28
(first lion can be put in any one of 8 cages & the second one can be put in any one of the remaining 7 cages; divide by 2 because the lions are identical & order is immaterial)
b. total ways that the lions can be put in adjacent cages (= ways to chose two adjacent cages) = 7
the answer = A - B = 21
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by diebeatsthegmat » Mon Nov 08, 2010 8:06 pm
agni_h1 wrote:Q.In how many ways can two identical lions can be put in eight identical cages such that the two lions are not in adjacent cages ?

No answer choices were given. I got the answer as 21 by taking one lion, one box and another lion as one unit. And then went on increasing the number of boxes between them.
is the answer you got OA?
because there are 2 lion so there must be 2 times/combination of choices
thus i think 21*2=42
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by Bharat » Mon Nov 08, 2010 9:35 pm
diebeatsthegmat wrote:
agni_h1 wrote:Q.In how many ways can two identical lions can be put in eight identical cages such that the two lions are not in adjacent cages ?

No answer choices were given. I got the answer as 21 by taking one lion, one box and another lion as one unit. And then went on increasing the number of boxes between them.
is the answer you got OA?
because there are 2 lion so there must be 2 times/combination of choices
thus i think 21*2=42
Please note that the lions are identical! hence order does not matter -> no need to multiply by two.

Thanks.
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by saurabhmahajan » Mon Nov 08, 2010 10:38 pm
I too got 42.
Let the lion be L1 & L2
for L1 total 8 choices
for L2 7 choices
So total choices 56 (8*7)
But they can sit adjacent so combining them will leave only 7 choices and as they can be L1-L2 or L2-L1 so total combined choices will be 14.
Now 56-14 = 42
Thanks and regards,
Saurabh Mahajan

I can understand you not winning,but i will not forgive you for not trying.
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by beat_gmat_09 » Mon Nov 08, 2010 10:49 pm
To solve this, get the number of ways the lions will be kept adjacent and subtract this from the total number of ways to arrange the identical lions.
As the lions are identical use combinations to get the total number of ways to arrange the lions.
Number of ways = C(8,2) = 28. You can also use the ANAGRAM method (manhattan).
LLCCCCCC, number of ways to arrange this is total number of ways/(no of ways lions)(no of ways cages)
= 8!/(2!X6!) = 28
Now get the number of ways in which the two lions are adjacent. Use logic to find the number of ways.
This can also be found out as = (n-1) = no of spaces - 1. In this case 8-1 = 7.
Diagrammatically (try out to get adjacent)-
_ _ _ _ _ _ _ _ There are only 7 possible ways to get adjacent cells.

Therefore total number of ways to arrange 2 lions in 8 cells such that lions are not adjacent =
28 - 7 = 21.
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