ME and THIS LOCK

Probability Level 2

I have a 4 digit lock for my locker. One day, I changed the code and accidentally 5 minutes later I forgot the code. All I ever knew is that I never used 2 same digits for the code. For example : I never used 5504 because there are 2 5's there. When I tried to open the lock, what is the probability I will succeed?? Express $\frac{a}{b}$ as $a+b$ .

5150 5041 5051 5040 5151 5140 5141 5050

2 solutions

Anwar Arif
Feb 22, 2015

(1/10) (1/9) (1/8)*(1/7)=1/5040 so the answer=5041

Richard Christian
Feb 11, 2015

As there are $4$ possible spaces for each $9$ digits, then we may count the first digit as $9$ digits available. If for example $1$ is the digit used in the first slot, then there are the digits $2, 3, 4, 5, 6, 7, 8,$ and $9$ available for the second slots, making ( $8$ ) digits available. In the 3rd slot, there may only be $7$ available digits, and $6$ for the fourth, so we can multiply $=$

$1$ (First digit) $* 8 * 7 * 6$ $=$ $504$

But, there are 10 possible digits for the first slot. So, we multiply $=$

$504 * 10$ $=$ $5040$

So, the probability of me getting it right is $\frac{1}{5040}$ . In the question, we must express $\frac{a}{b}$ as $a+b$ . So, we add $=$

$5040+1$ $=$ $\boxed{5041}$

ME and THIS LOCK

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