Find the number of ways in which 5 boys and 5 girls may be seated in a

But...
Gbgbgbbgbg

Gbbgbgbgbg

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IMO D
Boys 5! and girl 5!
120*120 = 14440
alternative sitting boys & girls = 14440*2 ; 28880 IMO E

I don't understand one thing. In the answer explanations, all the people assume that either first position was held by male
or by female, then female then male and then female in this way...But What will happen if two boys seat in consecutive position but no two girls seat together? The question did not mention that two boys cannot sit together. If we assume that two boys can also sit together in any place of the row, then won't the answer be changed? Pls someone explain where i was wrong.

If that happens, two girls have to sit together: BBGBGBGBGG

If we take a boy then for next sit will take a girl (as the condition says)

B X G X B X G X B X G X B X G X B X G
5 x 5 x 4 x 4 x 3 x 3 x 2 x 2 x 1 x 1 = 14400 (When starting form Boy first)

G X B X G X B X G X B X G X B X G X B
5 x 5 x 4 x 4 x 3 x 3 x 2 x 2 x 1 x 1 = 14400 (When starting form GRIL first)

TOTAL 28,800 WAYS
ANS. E

NO, NOT REALLY.
ONE OPTION CAN BE STARTING FROM GIRL FRIST GBGBGBGBGB
OTHER OPTION STARTING BY BOY FIRST BGBGBGBGBG

EITHER GIRL OR BOY IS NOT SEATING SIDE BY SIDE.

i had an issue with this question too....


Problem Type 1: if the question had said NO 2 BOYS --- AND ---- NO 2 GIRLS can sit together, then I agree, you have 2 Scenarios:

Scenario 1: Boys Fill the ODD Seats [1 - 3 - 5 - 7 - 9] and Girls fill the EVEN Seats

OR

Scenario 2: Boys Fill the EVEN Seats [2 - 4 - 6 - 8 - 10] and Girls fill the ODD Seats

for each Scenario, we can arrange the Boys in 5! ways and Arrange the Girls in 5! ways

2 * 5! * 5! = 28, 800 ---- Answer E





Problem Type 2 --- THIS QUESTION: It does NOT Say that the Boys must be separated. Boys are allowed to sit together. This is the way I answered this question:


1st) Sit the Boys 1st. We can Arrange the 5 Boys in 5! Ways.



2nd)Seat the 5 Boys with "GAPS" around Each Seat

___ A___B___C___D___E___


Now, based on the constraints in this Q-Stem, in order to sit the Girls such that NO 2 Girls are next to each other:

1st) We must Choose 5 out of the 6 Available "GAPS" in and around the Boys such that NO 2 Girls sit together

"6 Choose 5" = 6 Ways


1 Example Arrangement is the following:

if the Girls are: 1 --- 2 --- 3 --- 4 --- 5
the Boys are: A --- B --- C --- D ---- E


One possibility is:

1 - A - B - 2 - C - 3 - D - 4 - E - 5

or

g - B - B - g - B - g - B - g - B - g


This is why we must FIRST Choose the 5 Places where we are going to sit the Girls out of the 6 "GAPS" in and around the 5 Seated Boys

Again ---- "6 Choose 5" = 6 Ways to Choose



AND


2nd) Once we have the 5 Placements for the Girls, we can Arrange the Girls in 5! Ways.


Solution:

5! * "6 choose 5" * 5! = 5! * 6 * 5! = 120 * 720 = 86, 400 (assuming my math is correct)


The answer to this Question is assuming that the 5 Boys can NOT Sit together ALSO. However, the Q-Stem does NOT say this.


Any agreement? Am I wrong?

I think the answer is wrong for this answer. Lets start with siting with boys only.

Boy can be sited in 5! ways. one boys got sit, now we have 6 places for the girls to sit. But since we have only 5 girls so we need to select 5 position out of 6. This can be done in 6C5 ways i.e. 6 ways.

So total ways is 6C5 * 5! * 5! = 6 * 5! * 5!.

People are talking about 2 arrangements is possible is only when asked to sit in alternate only.

_ B1_B2_B3_B4_B5_ (6 dash for girls to sit).

Fdambro294 You're right. People have been assuming that there are only two arrangements (BGBGBGBGBG OR GBGBGBGBGB). However, the question just says that the girls may not sit next to each other, and they're sitting in a row (if the question said circle, this wouldn't be the case). There are really SIX arrangements:
alternating:
1. BGBGBGBGBG (girls in even chairs)
2. GBGBGBGBGB (girls in odd chairs)
girls as first and last, two boys clumped in the middle:
3. GBGBGBG(BB)G
4. GBGBG(BB)GBG
5. GBG(BB)GBGBG
6. G(BB)GBGBGBG

You can look at these arrangements as GBGBGBGBG (9 people fixed) with 6 different spots where the last boy can sit -- first, last, or next to each of the other 4 boys.

Therefore, 6 ways to choose B vs G seats, and 5! ways to order the boys and girls each, so the result is: 6*5!*5! = 86,400.
My student and I were unhappy with the official answer during tutoring.

Bunuel Please edit the OA to reflect the discrepancy, or fix the question to indicate that two boys also cannot sit next to each other.

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