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A group of 4 boys and 3 girls is to be seated in a row.

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A group of 4 boys and 3 girls is to be seated in a row.  [#permalink]

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New post 28 Mar 2018, 12:53
10
00:00
A
B
C
D
E

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  85% (hard)

Question Stats:

41% (01:51) correct 59% (02:04) wrong based on 76 sessions

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A group of 4 boys and 3 girls is to be seated in a row. How many such arrangements are possible where no 2 girls sit together?

A.144
B.288
C.576
D.720
E.1440
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A group of 4 boys and 3 girls is to be seated in a row.  [#permalink]

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New post 28 Mar 2018, 13:34
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prashants412 wrote:
A group of 4 boys and 3 girls is to be seated in a row. How many such arrangements are possible where no 2 girls sit together?

A.144
B.288
C.576
D.720
E.1440


If two girls cannot sit together, the group can sit together as follows
_ B _ B _ B _ B _

The three girls can be seated in either of the 5 marked spaces.
The total ways of placing the girls are \(C_3^5 = 10\)

There are \(24(4!)\) ways of arranging the boys and \(6(3!)\) ways of arranging the girls.

Therefore, the total arrangements possible when no 2 girls sit together are 6*24*10 = 1440 (Option E)
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Re: A group of 4 boys and 3 girls is to be seated in a row.  [#permalink]

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New post 28 Mar 2018, 17:03
1
Backing out unwanted arrangements approach:

Number of ways to arrange 7 people in a row: 7!=5040
Let's find the number of ways 2 or 3 girls can sit together and subtract that from all possible arrangements. Girls can sit together as follows:

1st type of arrangement
B G G B
Imagine a block of 4 people, with 2 girls sandwiched in between 2 boys.
Number of ways to arrange 2 girls for this block of 4 people: 3P2 = 6
Number of ways to arrange 2 boys for this block of 4 people: 4P2 = 12
If you think of the block of 4 people as 1 entity, and since we have 3 remaining spots, then the number of ways we can arrange this 1 block and the 3 remaining spots is the same as arranging 4 people: 4! = 24
Total arrangements: 6*12*24 = 1728

2nd type of arrangement
G G B _ _ _ _
This will occur when 2 girls occupy the first and second spots and a boy occupies the third spot.
Number of ways to arrange 2 girls: 3P2 = 6
Number of ways to arrange 1 boy: 4P1 = 4
Number of ways to arrange the remaining 4 people: 4! = 24
Total arrangements: 6*4*24=576

3rd type of arrangement
_ _ _ _ B G G
This will occur when 2 girls occupy the sixth and seventh spots and a boy occupies the fifth spot.
Same logic as for GGB.
Total arrangements: 576

4th type of arrangement
G G G
Imagine a block of 3 girls.
Number of ways to arrange 3 girls: 3P3 = 6
If you consider this block of 3 girls as 1 entity and since we have 4 other spots to fill, then we essentially have to arrange 5 entities. Number of arrangements: 5! = 120
Total arrangements: 6*120=720

So the number of unwanted arrangements is: 1728+576+576+720=3600

Total arrangements - unwanted arrangements = 5040-3600=1440

Answer: E
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Re: A group of 4 boys and 3 girls is to be seated in a row.  [#permalink]

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New post 28 Mar 2018, 21:25
prashants412 wrote:
A group of 4 boys and 3 girls is to be seated in a row. How many such arrangements are possible where no 2 girls sit together?

A.144
B.288
C.576
D.720
E.1440

Leftmost arrangement of constraint elements: g_g_ g _ _

By using the formula for finding the number of permutations , the number is 3!*4!((3+2+1) + (2+1) + 1))=144*10=1440

For rationale and explanation of the formula kindly check the link below
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A group of 4 boys and 3 girls is to be seated in a row.  [#permalink]

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New post 28 Mar 2018, 22:47
1
prashants412 wrote:
A group of 4 boys and 3 girls is to be seated in a row. How many such arrangements are possible where no 2 girls sit together?

A.144
B.288
C.576
D.720
E.1440



Most logical and Correct approach would be as shown by Pushpit..

Where do you have restrictions- on 2 girls sitting together.
So first let the boys sit in chairs with a gap in between..
_b_b_b_b_
GIRLS- So there are 5 gaps in which to place three girls, so 5C3= \(\frac{5!}{3!2!}=10\) but 3 girls can be arranged in 3!ways..
So 10*3!=60
BOYS- Butfor each arrangement of these 60, 4 boys can interchange places and can be arranged in 4! Ways..

Total =60*4!=1440


But say the restrictions were on 2 boys sitting together..
Place the 3girl...
_g_g_g_
So for boys there are 4 seats and 4 of them and you can choose 4out of 4 in 4C4=1
Ans 3!*4!*1
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Re: A group of 4 boys and 3 girls is to be seated in a row.  [#permalink]

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New post 28 Mar 2018, 23:19
2
1
in these type of problems always fix any one group.
Lets say i fix boys
so 4boys can be arranged in 4! ways.
now ,
-B-B-B-B-
the - marks the positions where girls can sit.
so first girl can sit in 5 ways
2nd girl in 4 ways
3 girl in 3 ways
hence my answer 4!*5*4*3 = 1440 (E)
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A group of 4 boys and 3 girls is to be seated in a row.  [#permalink]

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New post 03 Sep 2018, 08:15
chetan2u wrote:
prashants412 wrote:
A group of 4 boys and 3 girls is to be seated in a row. How many such arrangements are possible where no 2 girls sit together?

A.144
B.288
C.576
D.720
E.1440



Most logical and Correct approach would be as shown by Pushpit..

Where do you have restrictions- on 2 girls sitting together.
So first let the boys sit in chairs with a gap in between..
_b_b_b_b_
GIRLS- So there are 5 gaps in which to place three girls, so 5C3= \(\frac{5!}{3!2!}=10\) but 3 girls can be arranged in 3!ways..
So 10*3!=60
BOYS- Butfor each arrangement of these 60, 4 boys can interchange places and can be arranged in 4! Ways..

Total =60*4!=1440


But say the restrictions were on 2 boys sitting together..
Place the 3girl...
_g_g_g_
So for boys there are 4 seats and 4 of them and you can choose 4out of 4 in 4C4=1
Ans 3!*4!*1


Hi chetan2u

B_B_B_B this is ok ,
How about _BB_B_B here also we place Girls in the dashes then no two girls will be together.
again the group of two boys can move around so we can have _B_BB_B or _B_B_BB.

Does the solution given up till now cover all these cases? Thank you.
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Re: A group of 4 boys and 3 girls is to be seated in a row.  [#permalink]

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New post 03 Sep 2018, 08:50
1
stne wrote:
chetan2u wrote:
prashants412 wrote:
A group of 4 boys and 3 girls is to be seated in a row. How many such arrangements are possible where no 2 girls sit together?

A.144
B.288
C.576
D.720
E.1440



Most logical and Correct approach would be as shown by Pushpit..

Where do you have restrictions- on 2 girls sitting together.
So first let the boys sit in chairs with a gap in between..
_b_b_b_b_
GIRLS- So there are 5 gaps in which to place three girls, so 5C3= \(\frac{5!}{3!2!}=10\) but 3 girls can be arranged in 3!ways..
So 10*3!=60
BOYS- Butfor each arrangement of these 60, 4 boys can interchange places and can be arranged in 4! Ways..

Total =60*4!=1440


But say the restrictions were on 2 boys sitting together..
Place the 3girl...
_g_g_g_
So for boys there are 4 seats and 4 of them and you can choose 4out of 4 in 4C4=1
Ans 3!*4!*1


Hi chetan2u

B_B_B_B this is ok ,
How about _BB_B_B here also we place Girls in the dashes then no two girls will be together.
again the group of two boys can move around so we can have _B_BB_B or _B_B_BB.

Does the solution given up till now cover all these cases? Thank you.


Yes it covers all the cases..
Say we number the gaps..
1B2B3B4B5..
So we are choosing 3 out of these 5, say we choose 1,4 and 5
So GBBBGBG
Thus we cover all the cases
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A group of 4 boys and 3 girls is to be seated in a row.  [#permalink]

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New post 03 Sep 2018, 09:07
chetan2u wrote:
stne wrote:

Hi chetan2u

B_B_B_B this is ok ,
How about _BB_B_B here also we place Girls in the dashes then no two girls will be together.
again the group of two boys can move around so we can have _B_BB_B or _B_B_BB.

Does the solution given up till now cover all these cases? Thank you.


Yes it covers all the cases..
Say we number the gaps..
1B2B3B4B5..
So we are choosing 3 out of these 5, say we choose 1,4 and 5
So GBBBGBG
Thus we cover all the cases


Ohh.... I see now , I think I got it .
We were selecting the maximum number of positions available for the Girls, which is 5 , and then, when we choose 3 out of these 5 , we cover all the cases . Including BB_B_B_ types,and _BBB_B_ types let me know if I have understood it properly. Thanks .+1.
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Re: A group of 4 boys and 3 girls is to be seated in a row.  [#permalink]

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New post 03 Sep 2018, 09:24
stne wrote:
chetan2u wrote:
stne wrote:

Hi chetan2u

B_B_B_B this is ok ,
How about _BB_B_B here also we place Girls in the dashes then no two girls will be together.
again the group of two boys can move around so we can have _B_BB_B or _B_B_BB.

Does the solution given up till now cover all these cases? Thank you.


Yes it covers all the cases..
Say we number the gaps..
1B2B3B4B5..
So we are choosing 3 out of these 5, say we choose 1,4 and 5
So GBBBGBG
Thus we cover all the cases


Ohh.... I see now , I think I got it .
We were selecting the maximum number of positions available for the Girls, which is 5 , and then, when we choose 3 out of these 5 , we cover all the cases . Including BB_B_B_ types,and _BBB_B_ types let me know if I have understood it properly. Thanks .+1.



Yes you are absolutely correct in your understanding..
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Re: A group of 4 boys and 3 girls is to be seated in a row.  [#permalink]

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Re: A group of 4 boys and 3 girls is to be seated in a row.   [#permalink] 16 Sep 2019, 14:06
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