Solution Found!
(Michigan Mathematics Prize Competition, 1992, Part II)
Chapter 2, Problem 16E(choose chapter or problem)
(Michigan Mathematics Prize Competition, 1992, Part II) From the set \(\{1,2,3, \ldots, n\}\), \(k\) distinct integers are selected at random and arranged in numerical order (from lowest to highest). Let \(P(i, r, k, n)\) denote the probability that integer \(i\) is in position \(r\). For example, observe that \(P(1,2, k, n)=0\), as it is impossible for the number 1 to be in the second position after ordering.
(a) Compute \(P(2,1,6,10)\).
(b) Find a general formula for \(P(i, r, k, n)\).
Questions & Answers
QUESTION:
(Michigan Mathematics Prize Competition, 1992, Part II) From the set \(\{1,2,3, \ldots, n\}\), \(k\) distinct integers are selected at random and arranged in numerical order (from lowest to highest). Let \(P(i, r, k, n)\) denote the probability that integer \(i\) is in position \(r\). For example, observe that \(P(1,2, k, n)=0\), as it is impossible for the number 1 to be in the second position after ordering.
(a) Compute \(P(2,1,6,10)\).
(b) Find a general formula for \(P(i, r, k, n)\).
ANSWER:
Step 1 of 2
If the integer 2 is in the first position then the integer one should not be selected this one can be done in any one of \(1 C_{0}\ ways.The number 2 should be selected,which is done in \(1 C_{1}\). And the rest 6-1 numbers should be selected from 10-2 numbers. This can be done of any one of \(8 C_{5}\) ways . By the multiplication principle the product \(\left(1 C_{0}\right)\left(1 C_{1}\right)\left(8 C_{5}\right)\) equals the number of ways of the joint operation can be performed.
If we assume that each \(10 C_{6}\) of the ways of selecting 6 objects from the 10 objects.
Then we have to compute \(\mathrm{P}(2,1,6,10)\).
\(\mathrm{P}(2,1,6,10)=\frac{\left(1 C_{0}\right)\left(1 C_{1}\right)\left(8 C_{5}\right)}{10 C_{6}}\)
\(\mathrm{P}(2,1,6,10)=\frac{(1)(1)(56)}{210}\)
\(\mathrm{P}(2,1,6,10)=\frac{4}{15}\)
\(\text { Therefore } P(2,1,6,10)=\frac{4}{15}\) or 0.267