to earn a joint degree, a student must take any two maths classes from eight qths classes that are offered and any two physics classes from nine physics classes that are offered. How many different combination if he wishes to get a joint degree?
A) 68
B) 288
C) 988
D) 1008
E) 4032
Please help me solve this
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Hi PiyaNca,
To start, you would likely receive more of a response if you posted your question in the appropriate sub-forum. For example, the Problem Solving Forum can be found here:
https://www.beatthegmat.com/problem-solving-f6.html
In this question, we're told that to earn a joint degree, a student must take any two maths classes (from 8 classes that are offered) and any two physics classes (from 9 physics classes that are offered). We're asked for the number of different combinations of classes to get a joint degree.
Since we're dealing with COMBINATIONS of classes, the 'order' of the classes does NOT matter. We can use the Combination Formula to answer this question.
N!/K!(N-K)! where N is the total number of classes and K is the number in the sub-group.
For the 8 math classes, there are 8!/2!6! = (8)(7)/(2)(1) = 56/2 = 28 possible groups of 2 classes
For the 9 physics classes, there are 9!/2!7! = (9)(8)/(2)(1) = 72/2 = 36 possible pairs of 2 classes
Thus, the total number of combinations is (28)(36) = 1008
Final Answer: D
GMAT assassins aren't born, they're made,
Rich
To start, you would likely receive more of a response if you posted your question in the appropriate sub-forum. For example, the Problem Solving Forum can be found here:
https://www.beatthegmat.com/problem-solving-f6.html
In this question, we're told that to earn a joint degree, a student must take any two maths classes (from 8 classes that are offered) and any two physics classes (from 9 physics classes that are offered). We're asked for the number of different combinations of classes to get a joint degree.
Since we're dealing with COMBINATIONS of classes, the 'order' of the classes does NOT matter. We can use the Combination Formula to answer this question.
N!/K!(N-K)! where N is the total number of classes and K is the number in the sub-group.
For the 8 math classes, there are 8!/2!6! = (8)(7)/(2)(1) = 56/2 = 28 possible groups of 2 classes
For the 9 physics classes, there are 9!/2!7! = (9)(8)/(2)(1) = 72/2 = 36 possible pairs of 2 classes
Thus, the total number of combinations is (28)(36) = 1008
Final Answer: D
GMAT assassins aren't born, they're made,
Rich