Here's a common Permuatations and Combinations Questions:
a) The producers of a play require a total cast of 5, of which 3 are actors and 2 are actresses. He auditions 5 actors and 4 actresses for the cast. Find the total number of ways in which the cast can be obtained.
b) Find how many different odd 4-digit numbers LESS than 4000 can be made from the digits 1,2,3,4,5,6,7 if no digit may be repeated.
The FIRST thing to consider when solving Permutations and Combinations questions is recognizign what the hell you are doing...
ALWAYS REMEMBER:
PERMUTATIONS HAVE TO DO WITH ARRANGEMENTS OF THINGS
COMBINATIONS HAVE TO DO WITH HOW MANY POSSIBILITIES ARE THERE, WITHOUT WORRYING ABOUT ARRANGEMENTSif we follow that rule...
If for example, we had three things.... A,B,and C....
the combination ABC would be the same as BAC or CAB or BAC etc.
However the permutations are different...
Back to the question...
a) The producers of a play require a total cast of 5, of which 3 are actors and 2 are actresses. He auditions 5 actors and 4 actresses for the cast. Find the total number of ways in which the cast can be obtained.
This is obviously worried about combinations, not arrangements...
This my method of solving this question:
From: 5 actors 4 actresses
Choose: 3 actors 2 actresses
=> 5C3 x 4C2 = 10 x 6 = 60 combinations
b) Find how many different odd 4-digit numbers LESS than 4000 can be made from the digits 1,2,3,4,5,6,7 if no digit may be repeated.
Obviously this is a permutation question
its got to with arrangements....
This is my method in solving this;
First, find out how many
4-digit odd numbers you can have...
well
_ x _ x _ x _
We have 4 digits to fill from 7 numbers.... however for it to be an odd number, the last digit NEEDS to be either 1,3,5 or 7...
and that gives us four digits on the last one....
_ x _ x _ x
4What about the rest.... well.....
Obviously you will only have 6 choices for your first one, 5 for the next, and 4 for the next...
which gives us
6 x 5 x 4 x 4 = 480 odd numbers....
To find out how numbers from those odd numbers are
LESS than 4000...THEY NEED TO START WITH A 1, 2 or a 3... . and end with a 1,3,5 or 7...
First lets see all possibilites starting with 2....
Obviously because its starting with 2, we only have 1 option for the start.
1 x _ x _ x _
once again, we have 1,3,5,7 for the last option.... that gives us ... 4
1 x _ x _ x 4
how many numbers do we have left... 5 to choose from...
therefore.... 1 x 5 x 4 x 4 = 80 odd numbers Starting with 2, and that are less than 4000....
Now for numbers starting with 1 or 3....
we have two options for the first one....
2 x _ x _ x _
and we only have 3 options for the last one.... 1,3,5,7 but because we are already using one of them as our first number, we can only choose 3....
2 x _ x _ x 3
how many numbers do we have left? well we have.... 5...
therefore...
2 x 5 x 4 x 3 = 120....
120 + 80 = 200
We have 200, four digit odd numbers less than 4000...
Question 5 from October/November 2002 Paper 2...Cambridge
