Tresor Mukasa
asked on
Math 311 Discrete Math
In a class of 40 students, everyone has either a pierced nose or a pierced ear. The professor asks everyone with a pierced nose to raise his or her hand. Nine hands go up. Then the professor asked everyone with a pierced ear to do likewise. This time there are 33 hands raised. How many students have piercings both on their ears and their noses?
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aburr
answered
There is no unique answer. The answer
2 ~ Both
31 ~ ear(s) only
7 ~ nose only
0 ~ none
Is only one of 8 possible answers. (Provided there are no fractional students!)
The are 4 exhaustive and mutually exclusive categories, None ; Nose only ; Ears only ; & Both.
The 4 categories must sum to 40
The 2nd and 4th categories must sum to 9 (Nose)
The 3rd and 4th categories must sum to 33 (Ears)
Starting out with all students having a piercing (None category of zero, we can get a succession of valid solutions by changing by 1 each time.
So as Ears only decreases Both must increase [to maintain a total of 33 for Ears]. But this means Nose only must decrease [to maintain a total of 9 for Nose]. And finally the None category must increase[to maintain a total of 40 students].
None Nose only Ear(s) only Both Nose Ear(s)
0 7 31 2 9 33
1 6 30 3 9 33
2 5 29 4 9 33
3 4 28 5 9 33
4 3 27 6 9 33
5 2 26 7 9 33
6 1 25 8 9 33
7 0 24 9 9 33
((If we allow fractional students there are an infinite number of solutions between the first and last in the table above. However, I fear, there may be blood on the floor!))
Ian
2 ~ Both
31 ~ ear(s) only
7 ~ nose only
0 ~ none
Is only one of 8 possible answers. (Provided there are no fractional students!)
The are 4 exhaustive and mutually exclusive categories, None ; Nose only ; Ears only ; & Both.
The 4 categories must sum to 40
The 2nd and 4th categories must sum to 9 (Nose)
The 3rd and 4th categories must sum to 33 (Ears)
Starting out with all students having a piercing (None category of zero, we can get a succession of valid solutions by changing by 1 each time.
So as Ears only decreases Both must increase [to maintain a total of 33 for Ears]. But this means Nose only must decrease [to maintain a total of 9 for Nose]. And finally the None category must increase[to maintain a total of 40 students].
None Nose only Ear(s) only Both Nose Ear(s)
0 7 31 2 9 33
1 6 30 3 9 33
2 5 29 4 9 33
3 4 28 5 9 33
4 3 27 6 9 33
5 2 26 7 9 33
6 1 25 8 9 33
7 0 24 9 9 33
((If we allow fractional students there are an infinite number of solutions between the first and last in the table above. However, I fear, there may be blood on the floor!))
Ian
It saves time to read the question carefully before answering. The first sentence of the problem stipulates:
everyone has either a pierced nose or a pierced ear
So None = 0
And considering fractional students, fractional piercings, or students who are deaf, handless, or dishonest is obtuse rather than rigorous.
everyone has either a pierced nose or a pierced ear
So None = 0
And considering fractional students, fractional piercings, or students who are deaf, handless, or dishonest is obtuse rather than rigorous.
Hi there d-glitch,
Thanks for the pulling me up on this point. Bit of a red face all round here today!
Thanks for the pulling me up on this point. Bit of a red face all round here today!