What is the probability that the four digits are in sequence,either forward (suc

Question

What is the probability that the four digits are in sequence,either forward (such as 2345) or backward (such as 7654)? Note that there is no circling, so 9012 is not valid. I know you can count these up manually, but i can't seem tofigure out how to do this if say you had more than 4 digits and sayyou had 10 or something. Thank you. What is the probability that the four digits are in sequence,either forward (such as 2345) or backward (such as 7654)? Note that there is no circling, so 9012 is not valid. I know you can count these up manually, but i can't seem tofigure out how to do this if say you had more than 4 digits and sayyou had 10 or something. Thank you.

Explanation / Answer

First we have to find the no of ways we can arrange fournumbers from {0 , 1, ....,9} That is 10P4 Now we have to find the proper sequences The sequences are 0123.........6789 total 7 sequences arethere Since reverse seqence is also allowed, total no of sequencesare 14 So probability 14/10P4=1/360

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