solitaryreaper wrote:
Bunuel wrote:
SOLUTIONS:
1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?
A. 860/90,000
B. 810/100,000
C. 858/100,000
D. 860/100,000
E. 1530/100,000
Total # of 5 digit codes is 10^5, notice that it's not 9*10^4, since in a code we can have zero as the first digit.
# of passwords with three digit 6 is \(9*9*C^3_5=810\): each out of two other digits (not 6) has 9 choices, thus we have 9*9 and \(C^3_5\) is ways to choose which 3 digits will be 6's out of 5 digits we have.
\(P=\frac{favorable}{total}=\frac{810}{10^5}\)
Answer: B.
Hi Bunuel,
I have a doubt in 5C3 part in this problem.
I solved something like this:
Code will be something like this
6,6,6,A,B
this is like arranging 5 letters where 3 are exactly similar. It would give:
5! / 3! which is not equal to 5C3 .
Therefore I am getting a wrong answer.
Please correct my mistake (and why we are using Combination here while it looks like a case of Permutation).
Thanks in advance.
Now I am having a hard time to grasp how combination is giving the same answer. Can you please help me with the combination approach(how the problem translates to combination which is equivalent to the one the we got via permutation) ?
Moreover , how to decide when to use Permutation or Combination (and one that would lead to the solution in lesser time)
SR
Responding to a pm:
This is a question that haunts many - how do I know when to use permutation and when to use combination?
There is a simple solution - never use the permutation formula. The permutation formula finds very little direct use but leads to too many complications.
Always think in terms of selecting and arranging. For selecting r distinct elements out of n distinct elements, use nCr formula. Then arrange depending on whether the r elements selected need to be all distinct (r!) or some need to be same (a!/b!*c!) etc
Here, out of 9 digits you can select 2 in 9C2 ways and get a case such as 666AB. This will be arranged in 5!/3! ways.
Or you can select 1 digit out of 9 in 9C1 ways and get a case such as 666AA. This will be arranged in 5!/2!*3! ways.
9C2*5!/3! + 9C1*5!/2!*3! = 9*5!/3! * 9/2 = 810
Probability = 810/10^5
Note that you are using the Combinations formula only in this method too.
The other method using the combination formula just shows a different way of thinking.
You have 5 spots: _____ _____ _____ _____ _____
You choose any two spots out of these 5 in 5C2 ways.
For the first spot you choose, select a digit in 9 ways. For the second spot, select a digit in 9 ways.
In all remaining spots, just put 6.
This 5C2*9*9 directly gives you the total number of ways.
Note that 5C2 is the same as 5C3 (either you choose 2 spots for non 6 digits or you choose 3 spots for 6).
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