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# MTH100 - General Mathematics GDB No. 3 Solution Spring 2016 Due Date Jun 24, 2016

MTH100 - General Mathematics GDB No. 3 Solution Spring 2016 Due Date Jun 24, 2016

 Total Marks 5 Starting Date Tuesday, June 21, 2016 Closing Date Friday, June 24, 2016 Status Open Question Title Combination and Permutation Question Description Question:  Part a. Four vehicles are to be selected for a trip from a group of 6 jeeps and 4 cars. In How many different ways can they be selected such that at least one car should be there ?  Part b. How many 6 Digit password can be formed if the password cannot start with 0 and no digit appears more than one.    Grading criteria will be as follows: If formula of part a is correct then 25 % marks will be awarded.  If further process is correct then 50 % marks will  be awarded. If formula of part b is also correct  then 75% marks will be awarded.  If further process is correct then 100 % marks will  be awarded.                Note: It is mandatory to mention all necessary calculation steps to get full marks. Important Note:  If you need to type equation then use mathtype software and use the following link for the instructions to copy the math type equation on GDB forum

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Our main purpose here discussion not just Solution

We are here with you hands in hands to facilitate your learning and do not appreciate the idea of copying or replicating solutions.

Part a

Number of ways vehs can be grouped together

(3x Jeeps& 1 car) or (2x jeeps and 2 x cars) or (1x jeep and 3 x cars) or 4 cars

=(6C3×4C1)+(6C2×4C2)+(6C1×4C3)+(4C4)= 195

part b

=10P69P5
=1520015120=136080

Dear Students Don’t wait for solution post your problems here and discuss ... after discussion a perfect solution will come in a result. So, Start it now, replies here give your comments according to your knowledge and understandings....

part b (correction)

=10P6−9P5
=151200−15120=136080
part b (correction)

=10P6−9P5
=151200−15120=136080

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