We need to determinate who doesn't use anything, so first we must know the real number of who are practising: a person can do more things.
For doing this, is better use a three-circles graphic, but we try to do without it.
So
- 16 pool + gym + track, and this number is sure
- 38 gym + pool, but this number includes the people of 1st point. We must know who do only gym + pool: 38 - 16 = 22
- 31 pool + track, this number also includes people of 1st point, so: only pool + track 31 - 16 = 15
- 33 gym + track, this number also includes people of 1st point, so: only gym + track 33 - 16 = 17
now:
- 67 gym, we must know who use only gym: 67 - 16 (gym + pool + track) - 22 (gym + pool) - 17 (gym + track) = 12
- 62 pool, we must know who use only pool: 62 - 16 (gym + pool + track) - 22 (gym + pool) - 15 (pool + track) = 9
- 56 track, we must know who use only track: 56 - 16 (gym + pool + track) - 15 (pool + track) - 17 (gym + track) = 8
now we must now who do some facility: let's sum all
16 + 22 + 15 + 17 + 12 + 9 + 8 = 99
99/100 practise something
So only 1/100 doesn't use nothing
The probability is 1%
Answer:
5040,56
Step-by-step explanation:
We have to construct pass words of 4 digits
a) None of the digits can be repeated
We have total digits as 0 to 9.
4 digits can be selected form these 10 in 10P4 ways (since order matters in numbers)
No of passwords = 10P4
=
b) start with 5 and end in even digit
Here we restrain the choices by putting conditions
I digit is compulsorily 5 and hence only one way
Last digit can be any one of 0,2,4,6,8 hence 5 ways
Once first and last selected remaining 2 digits can be selected from remaining 8 digits in 8P2 ways (order counts here)
=56
The answer is 58 pretty sure
You might have to use parentheses to get 75...
The answer is
6(7+5) + 3
Need an explanation?