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.
Combination Problem
This topic has expert replies
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
Thanks.
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
Thanks.
-
- Legendary Member
- Posts: 1119
- Joined: Fri May 07, 2010 8:50 am
- Thanked: 29 times
- Followed by:3 members
is the answer you got OA?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.
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.diebeatsthegmat wrote:is the answer you got OA?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.
because there are 2 lion so there must be 2 times/combination of choices
thus i think 21*2=42
Thanks.
-
- Master | Next Rank: 500 Posts
- Posts: 250
- Joined: Thu Jan 07, 2010 2:21 am
- Thanked: 10 times
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
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.
Saurabh Mahajan
I can understand you not winning,but i will not forgive you for not trying.
-
- Master | Next Rank: 500 Posts
- Posts: 437
- Joined: Sat Nov 22, 2008 5:06 am
- Location: India
- Thanked: 50 times
- Followed by:1 members
- GMAT Score:580
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.
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.