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A group of 4 boys and 3 girls is to be seated in a row.
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28 Mar 2018, 11:53
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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.
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28 Mar 2018, 12:34
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.
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28 Mar 2018, 16:03
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 = 50403600=1440
Answer: E



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Re: A group of 4 boys and 3 girls is to be seated in a row.
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28 Mar 2018, 20:25
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.
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28 Mar 2018, 21:47
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.
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28 Mar 2018, 22:19
in these type of problems always fix any one group. Lets say i fix boys so 4boys can be arranged in 4! ways. now , BBBB 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.
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03 Sep 2018, 07:15
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 chetan2uB_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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 Stne



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Re: A group of 4 boys and 3 girls is to be seated in a row.
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03 Sep 2018, 07:50
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 chetan2uB_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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1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
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A group of 4 boys and 3 girls is to be seated in a row.
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03 Sep 2018, 08:07
chetan2u wrote: stne wrote: Hi chetan2uB_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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 Stne



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Re: A group of 4 boys and 3 girls is to be seated in a row.
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03 Sep 2018, 08:24
stne wrote: chetan2u wrote: stne wrote: Hi chetan2uB_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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1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
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Re: A group of 4 boys and 3 girls is to be seated in a row. &nbs
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