Permutation and Combination
- In a monthly test, the teacher decides that there will be there questions, one each from ex. 7, 8 and 9 of the text book. There are 12 questions in ex-7 , 18 in ex-8 and 9 in ex-9. In how many ways can the three questions be selected ?
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Number of ways selecting 1 question from ex-7 = 12C1
Number of ways selecting 1 question from ex-8 = 18C1
Number of ways selecting 1 question from ex-9 = 9C1Correct Option: A
Number of ways selecting 1 question from ex-7 = 12C1
Number of ways selecting 1 question from ex-8 = 18C1
Number of ways selecting 1 question from ex-9 = 9C1
∴ Total ways = 12C1 x 18C1 x 9C1 = 12 x 18 x 9 = 1944
- 139 person have signed for an elimination tournament. All players are to be paired up for the first round but because 139 is an odd number one player gets a bye, which promotes him to the second round without actually playing in the first round. The pairing continues on the next round with a bye to any player left over. If the schedule is planned, so that a minimum number of matches is required to determine the champion, the number of matches which must be played is ?
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Required number of member played will be (139 - 1) = 138
Correct Option: C
Required number of member played will be (139 - 1) = 138
- 2 Men and 1 women board a bus in which 5 seats are vacant, one of these 5 seats is reserved for ladies. A women may or may not sit on the seat reserved for ladies . In how many different ways can the five seats be occupied by these passengers ?
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Case I :-
If lady sets on reserved seat, then
2 men can occupy seats from 4 vacant seats in 4P2
= 4 x 3 = 12 ways
Case II :-
If lady does not site on reserved seat, then 1 women can occupy a seat from seat in 4 ways, 1 man can occupy a seat from 3 seats in 3 ways, also 1 man left can occupy a seat from remaining two seats in 2 ways.Correct Option: B
Case I :-
If lady sets on reserved seat, then
2 men can occupy seats from 4 vacant seats in 4P2
= 4 x 3 = 12 ways
Case II :-
If lady does not site on reserved seat, then 1 women can occupy a seat from seat in 4 ways, 1 man can occupy a seat from 3 seats in 3 ways, also 1 man left can occupy a seat from remaining two seats in 2 ways.
∴ Total ways = 4 x 3 x 2 = 24 ways
Hence, from Case I and case II , total ways = 12 + 24 = 36 ways
- Find the number of ways of arranging the host and 8 guests at a circular table, so that the host always sits in a particular seat ?
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Total number of persons = 9
Host can sit in a particular seat in one way .
Now, remaining positions are defined relative to the host .Correct Option: B
Total number of persons = 9
Host can sit in a particular seat in one way .
Now, remaining positions are defined relative to the host .
Hence, the remaining can sit in 8 places in 8P8 = 8! ways.
∴ The number of required arrangements = 8! x 1 = 8! = 8! ways
- How many different numbers can be formed from the digits 3, 4, 5, 6 and 7 when repetitions are allowed ?
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Number of 1 digit numbers = 5
Number of 2 digit numbers = 52 = 25
Number of 3 digit numbers = 53 = 125Correct Option: C
Number of 1 digit numbers = 5
Number of 2 digit numbers = 52 = 25
Number of 3 digit numbers = 53 = 125
Number of 4 digit numbers = 54 = 625
Number of 5 digit numbers = 55 = 3125
∴ Total number of numbers formed with these digits
= 5 + 25 + 125 + 625 + 3125 = 3905
