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\(B_1, B_2, B_3, B_4, B_5, G_1, G_2, G_3\) = 8 people

=> 8 people can be arranged in 8! ways: 40,320 ways

Girls sitting together: (\(G_1, G_2, G_3\)), \(B_1, B_2, B_3, B_4, B_5\) [Taking all girls as 1 set] = 6 people

=> 6 people can be arranged in 6! ways: 720 ways * 3 girls can be arranged in 3! ways = 720 * 6 = 4320

Overall ways for girls not to sit together: 40,320 - 4320 = 36000

Answer E
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The key phrasing is "ALL 3 Girls can not sit together."

If the Q-Stem had said "NONE of the 3 Girls can sit together," then it would be a Different Problem.


This means were are ONLY looking to eliminate any Arrangement that has ALL 3 GIRLS sitting next to each other.

G - G - B - G ------- would be OK!



(Total No. of Arrangements with NO RESTRICTION)
-
(No. of Arrangements in which ALL 3 Girls (G-G-G) sit Next to each other in VIOLATION of Restriction)

=

No. of Ways to seat the children so that ALL 3 Girls do NOT Sit Together


[8!] - [6! * 3!] =

[8 * 7 * (6!)] - [3 * 2 * (6!)] =

----take 6! Common----

(6!) * [56 - 6] =

6! * 50 = 36,000 Ways

-E-
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Hovkial
A teacher wishes to seat 5 boys and 3 girls in a row at a bench. In how many ways can the teacher seat the boys and the girls so that all the girls do not sit together?

(A) 720

(B) 2160

(C) 4032

(D) 18000

(E) 36000

Total ways of arranging 8 people in a row = 8!

Total ways in which all girls sit together (i.e. arrangement of 5 boys and group of girl which is 6! and arrangement among girls with-in group)= 6!*3!

Required outcomes = 8! - 6!3! = 6! (8*7-6) = 720*50 = 36000

Answer: Option E
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