 QUESTION

# A box contains coupons labeled 1,2,…..100. Five coupons are picked at random one after another without replacement. Let the number on the coupons be ${{x}_{1}}+{{x}_{2}}+.......+{{x}_{5}}$. What is the probability that ${{x}_{1}}>{{x}_{2}}>{{x}_{3}}$ and ${{x}_{3}}>{{x}_{4}}>{{x}_{5}}$ ?A. ${}^{1}/{}_{120}$B. ${}^{1}/{}_{60}$C. ${}^{1}/{}_{20}$D. ${}^{1}/{}_{10}$

Hint: First find the total number of cases that can be formed. ${{x}_{3}}$is smallest so keeping it fixed, find the cause of occurrence of ${{x}_{1}}>{{x}_{2}}>{{x}_{3}}$ and ${{x}_{3}}>{{x}_{4}}>{{x}_{5}}$. Thus find the probability by dividing it by total number of cases.

Given that a box contains coupons labeled 1,2,….,100, which means that there are a total of 100 coupons.

Now if we are selecting 5 coupons among this without replacement, then the number of ways we can pick up 5 coupons are 5! ways.

Thus the total number of cases of picking 5 random coupons from a box of 100 coupons are $^{100}{{C}_{5}}\times 5!$.

Thus we got total cases ${{=}^{100}}{{C}_{5}}\times 5!$.

Now we need to take favorable number of cases so ${{x}_{3}}$ is always the smallest among the 5 coupons, ${{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}},{{x}_{5}}$.

Thus ${{x}_{3}}$ is fixed as the smallest number. Now we have remaining 4 coupons out of 5 picked up.

Out of this we can pick 2 numbers from the 4 coupons in $^{4}{{C}_{2}}$ ways and we can assign them as ${{x}_{1}}$ and ${{x}_{2}}$.

Now there is only one way as the condition, ${{x}_{1}}>{{x}_{2}}$. Hence the rest of the 2 numbers are also fixed.

So the total number of favorable outcomes for ${{x}_{1}}>{{x}_{2}}>{{x}_{3}}$ and
${{x}_{3}}>{{x}_{4}}>{{x}_{5}}$ is $^{100}{{C}_{5}}{{\times }^{4}}{{C}_{2}}$.

Thus the probability of occurrence of ${{x}_{3}}>{{x}_{4}}>{{x}_{5}}$ and

${{x}_{1}}>{{x}_{2}}>{{x}_{3}}$ out of the total number of cases is:

$=\dfrac{^{100}{{C}_{5}}{{\times }^{4}}{{C}_{2}}}{^{100}{{C}_{5}}\times 5!}=\dfrac{^{4}{{C}_{2}}}{5!}$

We know that $^{n}{{C}_{r}}=\dfrac{n!}{(n-r)!r!}$.

\begin{align} & {{\therefore }^{4}}{{C}_{{}^{2}/{}_{5!}}}=\dfrac{4!}{(4-2)!2!5!}=\dfrac{4!}{2!2!5!}=\dfrac{4!}{2!2!5\times 4!} \\ & =\dfrac{1}{2\times 2\times 5}={}^{1}/{}_{4}\times 5={}^{1}/{}_{20}. \\ \end{align}

Thus we got the probability as ${}^{1}/{}_{20}$.

Option C is the correct answer.

Note: We have used combinations here because the coupons to be arranged are not to be arranged in order. It is not important to arrange them in order. Thus we use combination here and not permutation. Try to identify the important events involved in the given problem.