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29 Apr 2026, 19:00
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
(N/A)
Question Stats:
67% (02:24) correct
33%
(02:29)
wrong
based on 18
sessions
History
Date
Time
Result
Not Attempted Yet
Four boys and three girls are to be seated in a row. How many different arrangements are possible if two particular girls – Rita and Maria – must sit together and two particular boys – Sam and Tim – must not sit together?
A. 240
B. 720
C. 960
D. 1200
E. 1440
A. 240
B. 720
C. 960
D. 1200
E. 1440
29 Apr 2026, 22:02
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Take Rita and Maria together first:
Ways to arrange = 6!*2 = 1440.
Now we need to remove Sam and Tim together cases:
When R and M as well and S and T are together, ways to arrange:
= 5!*4 = 480.
So total cases = 1440 - 480 = 960.
Answer: Option C
Ways to arrange = 6!*2 = 1440.
Now we need to remove Sam and Tim together cases:
When R and M as well and S and T are together, ways to arrange:
= 5!*4 = 480.
So total cases = 1440 - 480 = 960.
Answer: Option C
ExpertsGlobal5
BongBideshini
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01 May 2026, 06:40
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Treat Rita & Maria as one block.
(RM), Sam, Tim, 2 other boys, 1 other girl = 6 units
Arrangements:
6! × 2 = 1440
(RM), (ST), 2 other boys, 1 girl = 5 units
Arrangements:
5! × 2 × 2 = 480
Required arrangements:
1440 - 480 = 960
Answer: C. 960
(RM), Sam, Tim, 2 other boys, 1 other girl = 6 units
Arrangements:
6! × 2 = 1440
(RM), (ST), 2 other boys, 1 girl = 5 units
Arrangements:
5! × 2 × 2 = 480
Required arrangements:
1440 - 480 = 960
Answer: C. 960
ExpertsGlobal5
