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17 Dec 2025, 10:24
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As per the question we have data of 4 people [i.e. Alex (A), Burno (B), Carlo (C), Denise (D)] We have 2 condition:
1st condition when remaining 2 people are between A,B,C,D: _A_B_C_D_ so we have 5 position to select 2 position 5C2 and they can interchange: 5C2 * 2! = 20.
2nd condition when they both are together: _ _ A _ _ B _ _ C _ _ D _ _ so we have a tie between 2 people with 5 position and they can interchange within 1 tie: 5*2= 10
1st condition + 2nd condition= 20+10=30 (D)
1st condition when remaining 2 people are between A,B,C,D: _A_B_C_D_ so we have 5 position to select 2 position 5C2 and they can interchange: 5C2 * 2! = 20.
2nd condition when they both are together: _ _ A _ _ B _ _ C _ _ D _ _ so we have a tie between 2 people with 5 position and they can interchange within 1 tie: 5*2= 10
1st condition + 2nd condition= 20+10=30 (D)
17 Dec 2025, 10:29
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A speaks before B, B before C, C before D
Total speaker =6
Let other 2 speakers be X & Y
Total way to arrange 6 speakers =6! =6*5*4*3*2*1=720
Way to arrange A,B,C,D wihtout order =4! =4*3*2*1=24
Out of possible 24 orders A,B,C,D can only be arranged in 1 way = 1/24 ways
Ways of scheduling 6 speakers = 720*1/24=30 ways
D
Total speaker =6
Let other 2 speakers be X & Y
Total way to arrange 6 speakers =6! =6*5*4*3*2*1=720
Way to arrange A,B,C,D wihtout order =4! =4*3*2*1=24
Out of possible 24 orders A,B,C,D can only be arranged in 1 way = 1/24 ways
Ways of scheduling 6 speakers = 720*1/24=30 ways
D
17 Dec 2025, 10:38
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Bunuel
We have 6 speakers total, with the required order among 4 of them:
A<B<C<D
Without restrictions (possible schedule): 6! = 720
Number of positive relative orders for A,B,C,D: 4!=24
Fraction that satisfy A<B<C<D: (1/4!)
Valid schedules: 6!/4!=30
OPTION D.
