a three digit number with no digit repeated is made from the digits 5,6,7,8 if you are asked to choose a number what is the probability that the number is a.odd b.even c.a multiple of 5 d.a number greater than 750
Answers
Answer:
Repetition allowed. 5,6,7,8,9,0 : Six digits.
The first place can be filled by 6,7,8,9 i.e. in 4 ways as the number is to be greater than 600000. The last place can be filled in by 5, 7, 9 i.e. 3 ways as the number is to be odd and because of repetition. Hence the number of ways of filling the 1st and the last place is
4×3=12 ways
We have to fill in the remaining 4 places of the six digit number i.e. 2nd, 3rd, 4th and 5th place. Since repetition is allowed each place can be filled in 6 ways. Hence the 4 places can be filled in 6×6×6×6=1296 ways.
Hence by fundamental theorem the total numbers will be
1296×12=15552.
Repetition not allowed.
1st Place Last (Number to be odd)
6 5,7,9=3 ways
8 5,7,9=3 ways
7 5,9=2 ways (not 7)
9 5,7=2 ways (not 9)
Total number of ways of filling the first and the last place under given condition is
3+3+2+2=10 ways.
(It was 12 when repetition was allowed).
Having filled in the first and the last places in 10 ways, we can fill in the remaining 4 places out of 4 digits (repetition not allowed) in 4! = 24 ways.
Total number of ways is 10×24=240
Answer:
Repetition allowed. 5,6,7,8,9,0 : Six digits.
The first place can be filled by 6,7,8,9 i.e. in 4 ways as the number is to be greater than 600000. The last place can be filled in by 5, 7, 9 i.e. 3 ways as the number is to be odd and because of repetition. Hence the number of ways of filling the 1st and the last place is
4×3=12 ways
We have to fill in the remaining 4 places of the six digit number i.e. 2nd, 3rd, 4th and 5th place. Since repetition is allowed each place can be filled in 6 ways. Hence the 4 places can be filled in 6×6×6×6=1296 ways.
Hence by fundamental theorem the total numbers will be
1296×12=15552.
Repetition not allowed.
1st Place Last (Number to be odd)
6 5,7,9=3 ways
8 5,7,9=3 ways
7 5,9=2 ways (not 7)
9 5,7=2 ways (not 9)
Total number of ways of filling the first and the last place under given condition is
3+3+2+2=10 ways.
(It was 12 when repetition was allowed).
Having filled in the first and the last places in 10 ways, we can fill in the remaining 4 places out of 4 digits (repetition not allowed) in 4! = 24 ways.
Total number of ways is 10×24=240
Explanation:
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