In how many ways can five girls stand in line if Maggie and Lisa canno

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



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

Total ways (w/o restriction) 5 girls can be arranged in 5! ways = 120

Total ways in which Maggie and Lisa cannot stand next to each other = Total ways - total ways in which Maggie and Lisa stand next to each other

= 120 - (4!*2) = 120 - 48 = 72 ways.

Ans. D, 72

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.

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).

Glue method (Maggie and Lisa stand together)

5!-(4!*2)=72

D

We can use the formula:

Number of ways Maggie is not next to Lisa = total number of arrangements - number of ways Maggie is next to Lisa

The total number of arrangements with no restrictions is 5! = 120.

Maggie is next to Lisa can be shown as:

[M-L] - A - B - C

Since Maggie and Lisa are now represented as one person, there are 4! ways to arrange the group and 2! ways to arrange Maggie and Lisa. Thus, we have 4! x 2! = 24 x 2 = 48 ways for Maggie to stand next to Lisa.

Thus, the number of ways to arrange Maggie and Lisa such that they are not together is 120 - 48 = 72.

Answer: D

Hi Friends,

Can u anybody tell me how to solve this problem I am just bit modify this problem!

In how many ways can five girls stand in line if Maggie, Lisa, and Jane cannot stand next to each other?

Hi,

your formulas, don't make any sense, you wrote them for n=k, please edit them. Thank you!

Leftmost arrangement of constraint elements is : M_L_ _

Using formula, we have number of permutations as 2!*3!*(3+2+1)=72

For explanation of the formula kindly see the link below.

M & L can be separated in 6 ways, in these positions:
1 & 3
1 & 4
1 & 5
2 & 4
2 & 5
3 & 5
each of these 6 ways allows 12 possibilities:
2 for M & L:
ML
LM
* 6 for the other 3 girls:
ABC
ACB
BAC
BCA
CAB
CBA
6*12=72
D

Hi Friends,

Can u anybody tell me how to solve this problem I am just bit modify this problem!

In how many ways can five girls stand in line if Maggie, Lisa, and Jane cannot stand next to each other?

Hello VaibNop , I have looked at your attempt. The same is ok to visualize but time consuming for exam.
This is a problem of combinatorics. Lets understand when only 2 of the 5 can not stand side by side . then we will apply the logic in your query 3 of 5.

Case 01: when only 2 of the 5 can not stand side by side

Now the total number of ways 5 people can be arranged =5P5=5!=120
The Q asked in how many cases 2 can not stand side by side.
Lets find the #cases where 2 can stand side by side . ( then, # ways when 2 donot stand side by side = total # ways 5 people stand - #cases where 2 can stand side by side)

Lets assume these 2 are actually 1 object , hence our modified total = 4 objects. Now the total number of ways 4 objects can be arranged =4P4=4!=24.
Now the 2 objects , which we considered to be a single object can be arranged among them selves in ways =2P2=2!=2 .
#cases where 2 can stand side by side = 24*2 = 48

Thus , # ways when 2 donot stand side by side = total # ways 5 people stand - #cases where 2 can stand side by side = 120 - 48 = 72 ways


Case 02: when only 3 of the 5 can not stand side by side ....................Please note how we are just plugging in the values to our earlier understanding.


Now the total number of ways 5 people can be arranged =5P5=5!=120
The Q asked in how many cases 3 can not stand side by side.Lets assume these 3 are actually 1 object , hence our modified total = 3 objects. Now the total number of ways 3 objects can be arranged =3P3=3!=6.
Now these 3 objects , which we considered to be a single object can be arranged among them selves in ways =3P3=3!=6 .
#cases where 3 can stand side by side = 6*6 = 36

Thus , # ways when 3 donot stand side by side = total # ways 5 people stand - #cases where 3 can stand side by side = 120 - 36 = 84 ways

total ways ; 5! = 120
and ML together ; 4! and adjacent 4!*2 = 48
so nor together ; 120-48 ; 72
IMO D

I solved in this way:

[2(MorL)x3(ABC)x3(LBC)x2(BC)x1(C)] + [1(C)x2(BC)x3(LBC)x3(ABC)x2(MorL)]

is it wrong?

Thx