Math, asked by username52270, 1 year ago

In how many ways u can arrange 5 lions and 4 tigers so that no two tigers follow each other .
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Answers

Answered by abs2001
1
1st Approach :

L T L T L T L T L

The 5 lions should be arranged in the 5 places marked 'L'

This can be done in 5! ways.

The 4 tigers should be in the 4 places marked 'T'

This can be done in 4! ways.

Therefore, the lions and the tigers can be arranged in 5! ×× 4! ways = 2880 ways.

2nd Approach :

But I want to approach in the following way :

Let all the lions appear together so that 4L = 1unit

So, there are 4 tigers + 1 Lion = 5 units

5 things can be arranged in 5!5! ways and 5 lions can be arranged themselves in 5! ways , So there are 5! ×5!×5! ways.

Total number of ways in which we can arrange 9 items = 9! ways.

Therefore condition when no two lions never appear together

= 9! - 5!×5!5!×5! = 5!( 9.8.7.6 - 5.4.3.2) = 348480



I hope this helps you!!!

Please mark my answer as the brainliest.


username52270: jadha bange sala
sanjay270899: username52270 can tell the right answer...
abs2001: First make tigers sit:
_T_T_T_T_

Now send the lions in the gaps:
LTLTLTLTL

Tigers can interchange: 4! Ways
Ways to make the Lions sit: 5C5 * 5! (5C5=1)

So, final answer: 4!*5!=120*24= 2880
abs2001: I hope this one is right
username52270: No
username52270: sanjay has given right answer
username52270: 43200
abs2001: ok then sorry
username52270: ok
abs2001: incovinience is highly regretted
Answered by Anonymous
2
hey, here is your answer.....


==> 43,200 ways...




hope this helps you...
thank you....


plzz mark me as brainliest .....
Attachments:

username52270: abs2001
sanjay270899: First we will keep all 5 lions in one line... Now all five lions can be arranged in 5! methods. Now there are 6 spaces between all 5 lions, so 4 tigers can be at any of these six spaces: 6C4 ways "here your above mention case will be consider". Now all four arranged tigers can be arrange in 4! ways. Thus answer is 5! 4! 6C4.
username52270: pls.try
username52270: thank u sanjay
Anonymous: hey sanjay, its not 6C4, its "6p4"...
sanjay270899: No, you haven't read my comment correctly...
sanjay270899: In your answer you have used "permutations" but I have used "combinations".
sanjay270899: By both methods our answer will be same... I generally prefer "Combination" method.
Anonymous: oh.. sorry, u r right
sanjay270899: It's Ok.
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