Question: 31 -
There are 8 orators A, B, C, D, E, F, G and H. In how many ways can the arrangements be made so that A always comes before B and B always comes before C.
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8! / (5! x 3!)
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8! / 6!
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5! x 3!
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8! / 3!
Answer:
8! / 3!
Solution:
Select any three places for A, B and C. They need no arrangement amongst themselves as A would always come before B and B would always come before C.
The remaining 5 people have to be arranged in 5 places.
Thus, 8C3 x 5! = 56 x 120 = 67200 = 8! / 3!
Select any three places for A, B and C. They need no arrangement amongst themselves as A would always come before B and B would always come before C.
The remaining 5 people have to be arranged in 5 places.
Thus, 8C3 x 5! = 56 x 120 = 67200 = 8! / 3!
Question: 32 -
From among the 36 students in a class, one leader and one class representative are to be appointed. In how many ways can this be done?
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1260
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1360
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1160
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1060
Answer:
1260
Solution:
There are 36 students and every one has equal chance of being selected as a leader. Hence, the leader can be appointed in 36 ways. When one person is appointed as leader, we are left with 35 students. Out of these 35 teachers, we can select one class representative. So, a class representative can be selected in 35 ways. Hence, the number of ways in which a leader and class representative can be selected = 36 x 35 = 1260
There are 36 students and every one has equal chance of being selected as a leader. Hence, the leader can be appointed in 36 ways. When one person is appointed as leader, we are left with 35 students. Out of these 35 teachers, we can select one class representative. So, a class representative can be selected in 35 ways. Hence, the number of ways in which a leader and class representative can be selected = 36 x 35 = 1260
Question: 33 -
How many 5 digit even numbers with distinct digits can be formed using the digits 1, 2, 5, 5, 4?
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24
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36
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16
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48
Answer:
24
Solution:
The 5 digit even numbers can be formed out of 1, 2, 5, 5, 4 by using either 2 or 4 in the unit’s place. This can be done in 2 ways.
Corresponding to each such arrangement, the remaining 4 places can be filled up by any of the remaining four digits in 4! / 2! = 12 ways. [5 is repeating twice hance 2! in denominator]
Hence, the total number of words = 2 x 12 = 24.
The 5 digit even numbers can be formed out of 1, 2, 5, 5, 4 by using either 2 or 4 in the unit’s place. This can be done in 2 ways.
Corresponding to each such arrangement, the remaining 4 places can be filled up by any of the remaining four digits in 4! / 2! = 12 ways. [5 is repeating twice hance 2! in denominator]
Hence, the total number of words = 2 x 12 = 24.
Question: 34 -
There are three rooms in a motel: one single, one double and one for four persons. How many ways are there to house seven persons in these rooms ?
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7! / 1! 2! 3!
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7!
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7! / 3
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7! / 3!
Answer:
7! / 1! 2! 3!
Solution:
Choose 1 person for the single room & from the remaining choose 2 for the double room & from the remaining choose 4 people for the four person room,
Then, 7C1 x 6C2 x 4C4 = 7! / 1! 2! 3!
Choose 1 person for the single room & from the remaining choose 2 for the double room & from the remaining choose 4 people for the four person room,
Then, 7C1 x 6C2 x 4C4 = 7! / 1! 2! 3!
Question: 35 -
There are 6 letters for 3 envelopes. In how many different ways can the envelopes be filled?
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110
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120
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130
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100
Answer:
120
Solution:
The 1st envelope can be filled up in 6 ways.
The 2nd envelope can be filled up in 5 ways and the 3rd envelope can be filled up in 4 ways.
Therefore, by the principle of association, the three envelopes can be filled up in 6 x 5 x 4 = 120 ways
The 1st envelope can be filled up in 6 ways.
The 2nd envelope can be filled up in 5 ways and the 3rd envelope can be filled up in 4 ways.
Therefore, by the principle of association, the three envelopes can be filled up in 6 x 5 x 4 = 120 ways