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Parik Parik
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Posts: 22
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9 years ago
31.   A pharmacist uses five separate weights 1g, 2g, 4g, 8, and 16g. If the pharmacist can combine these weights to create new weights, how different weights are possible? Show your calculations.

32.   At an alternative learning school, a group of 48 students enrolled in mathematics, French, and physics. Some students were more successful than others: 32 passed French; 27 passed physics; 33 passed mathematics; 26 passed French and mathematics; 26 passed physics and mathematics; 21 passed French and physics; and 21 passed French, mathematics, and physics. How many students passed one or more of the subjects? Solve the question using the inclusion-exclusion principle. Show your calculation.
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wrote...
9 years ago
31) The answer is 120, found by multiplying 5 x 4 x 3 x 2 x 1 ..

this is the result of a factorial.. written as 5! meaning 5x4x3x2x1

This covers the many different combinations possible using any number of weights. (Each one only once though as the pharmacist only has one of each)

32) (32 + 27 + 33) - (26 + 26 + 21) + 21 = 40 people passed at least one class.

(32 + 27 + 33) is the number of people that passed something (we are double/triple counting)

(26 + 26 + 21) is the number of people that passed two classes, exclude these (these are the people we counted twice in the previous step.)

21 people passed all 3 classes. include this (they were removed thrice in the previous step, so we add them again)
Parik Author
wrote...
9 years ago
31) The answer is 120, found by multiplying 5 x 4 x 3 x 2 x 1 ..

this is the result of a factorial.. written as 5! meaning 5x4x3x2x1

This covers the many different combinations possible using any number of weights. (Each one only once though as the pharmacist only has one of each)

32) (32 + 27 + 33) - (26 + 26 + 21) + 21 = 40 people passed at least one class.

(32 + 27 + 33) is the number of people that passed something (we are double/triple counting)

(26 + 26 + 21) is the number of people that passed two classes, exclude these (these are the people we counted twice in the previous step.)

21 people passed all 3 classes. include this (they were removed thrice in the previous step, so we add them again)


Hey biochikaa! Thank you for your answers!! I got the same answer for 32 so that seems right. But for # 31 I got a very different answer.


My answer for 31 is:

31.
1g, 2g, 4g, 8g, and 16g = 5 weights.
Number of weights with 1 combinations = C (5,1) = 5C1, (5¦1) = 5
Number of weights with 2 combinations = C (5,2) = 5C2, (5¦2) = 10
Number of weights with 3 combinations = C (5,3) = 5C3, (5¦3) = 10
Number of weights with 4 combinations = C (5,4) = 5C4, (5¦4) = 5
Number of weights with 5 combinations = C (5,5) = 5C5, (5¦5) = 1
Total number of combinations is 5+10+10+5+1 = 31. Therefore, there are 31 different possible weights. 
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