Problem 2 [22 pts (5, 4, 1, 3, 3, 6)]: Counting i. Students enrolled in CS1800 a

Question

Problem 2 [22 pts (5, 4, 1, 3, 3, 6)]: Counting i. Students enrolled in CS1800 are 300 women and 423 men. There are four sections, each of size at least 171, and each with at most 80 women enrolled. Prove that at least one section has at least 100 men, and that no section has more than 150 men enrolled. ii. A math professor wears for each class a combination of: one of his three hats (red, blue, green); one of his pants (black, blue, white, green); one of his shirts (white, red, black); and one of his pair of shoes (black, red, blue, white). He never wears combinations with three items of the same color. How many different combinations are wearable?

Explanation / Answer

(i)
300 women , 423 men
4 section each of size at least 171 and each with at most 80 women.
To prove: Atleast one section has at least 100 men.
Proof by contradiction: Let us suppose no section has more than 100 men. Now if every section has 99 men, total 4 sections will have 99 x 4 = 396 men. But we have to accomodate 423 men in four sections. Hence our assumption was wrong and at least one section will have at least 100 men.
To prove: No section has more than 150 men enrolled.
Proof by contradiction: Let there be four sections A, B, C and D
Section A : 151 men 20 women (Since every section has to have at least 171), Remaining men 423 - 151 = 272
Total students left = 272 men + 280 women = 552 to be divided into 3 sections.
Since each section can have atmost 80 women , we allocate 80 women to each Section B, C and D. Leftover women = 280 - (80 x 3) = 40. 40 women goes to Section A : 151 men and 60 women.
Now Section B + C+ D = 272 men + 240 women : Total 512. 512 /3 =170.67 < 171. Hence , if Section A has 151 men at least one of the Sections B, C and D will have less than 171 students which does not fall within the given constraint of having atleast 171 students in every section. Hence our assumption was wrong and no section can have more than 150 men.

(ii)
3 hats - red , blue and green
4 pants - black , blue , white , green
3 shirts - white , red , black
4 pair of shoes - black , red , blue , white
Total Combinations: 3 Hats x 4 Pants x 3 Shirts x 4 Shoes = 144
From 144 we subtract the combinations in which any 3 items are of the same colour.
144 - 4 (Red Hat, Red Shirt, Red Shoes, 4 X Pants) - 3 (Blue Hat, Blue Pants, Blue Shoes, 3 X Shirts) - 3 (Black Pants , Black Shirt , Black Shoes, 3 X Hat) - 3(White Pants, White Shirts, White Shoes, 3 X Hats) = 144 - 4 - 3 - 3 - 3 = 131 combinations are wearable

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