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Re: In how many ways can five girls stand in line if Maggie and Lisa canno [#permalink]
05 Feb 2010, 12:31
1
This post received KUDOS
Expert's post
1. the total number of permutations: 5! = 120 2. Let's consider Maggie and Lisa as one object, then the total number of permutations with Maggie and Lisa together: 4! = 24. 3. Take into account that [Maggie, Lisa] and [Lisa, Maggie] are different. 4. Maggie and Lisa cannot stand next to each other in: 120 - 2*24 = 72 ways. _________________
Re: In how many ways can five girls stand in line if Maggie and Lisa canno [#permalink]
26 Jul 2015, 23:09
can we solve this with \(P^n_r\) formula? _________________
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Re: In how many ways can five girls stand in line if Maggie and Lisa canno [#permalink]
27 Jul 2015, 04:42
GMATinsight wrote:
Subanta wrote:
can we solve this with \(P^n_r\) formula?
Jut a personal Suggestion: It's best to see the steps and work in steps in P&C problems rather than being dependent on formulas
However, Answer to your query is as follows
Quote:
In how many ways can five girls stand in line if Maggie and Lisa cannot stand next to each other?
(A) 112 (B) 96 (C) 84 (D) 72 (E) 60
5 girls can stand in line in \(P^5_5 = 120 ways\)
5 girls can stand in line in such that Maggie and Lisa stand together in \(P^4_4 * P^2_2= 48 ways\)
Total favourable ways of arrangement of five girls such that Maggie and Lisa cannot stand next to each other = 120 - 48 = 72 ways
Answer: option D
Thanks. I usually solve my problems without the formulae, but I have come across many problems where it is easier to use the formulae. I'm only trying to get familiar with the usage of the formulae. _________________
Kudos = Thanks
"Yeah, you can get a nickel for boosting Starfall, but jacking Heal's a ten-day stint in county. Now lifting Faerie Fire, they just let you go for that — it's not even worth the paperwork. But Reincarnation, man! That'll get you life!"
Re: In how many ways can five girls stand in line if Maggie and Lisa canno [#permalink]
27 Jul 2015, 04:46
Five girls can stand in a line in 5! = 120 ways.
Let M(Maggie) and L(Lisa) be treated as a single person ML. Now, ML can be placed in -x-x-x- any one of the 4 empty slots in 4! = 24 ways. ML can be ordered between themselves in 2! ways. So, the number of ways ML can stand together = 24 * 2 = 48.
So, the number of ways they don't stand together is 120 - 48 = 72 ways. Ans (D). _________________
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