Question

A box contains 2 fifty paise coins, 5 twenty five paisa coins and a certain fixed number N $\left( \ge 2 \right)$ of 10 and five paise coins. Five coins are taken out of the box at random. The number of combinations that the total value of these 5 coins is less than one rupee and fifty paisa is $10\left( N+k \right)$ . Find the value of k.

Answer 1

First, we will find total number of coins. Then we are given a number of combinations of 5 coins whose value is less than one rupee and fifty paisa. So, we will consider event ‘E’ whose total of 5 coins on selection is more than or equal to one rupee and fifty paisa. So, there will be three possibility such as First, taking one coin of fifty paise and 4 coins of twenty-five paisa $={}^{2}{{C}_{1}}\times {}^{5}{{C}_{4}}\times {}^{N}{{C}_{0}}$ . Second, taking two coins of fifty paise and 3 coins of twenty-five paisa $={}^{2}{{C}_{2}}\times {}^{5}{{C}_{3}}\times {}^{N}{{C}_{0}}$ .Third, taking two coins of fifty paisa, 3 coins of twenty-five paisa and one coin of 10 and five paisa $={}^{2}{{C}_{2}}\times {}^{5}{{C}_{2}}\times {}^{N}{{C}_{1}}$ . On adding all these and solving using the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ , we will get answer in form of $10\left( N+k \right)$ . Thus, then we will get the value of k on comparing.

Complete step-by-step answer:
Here, first we will calculate the total number of coins given in question. So, we have 2 fifty paise coins, 5 twenty five paisa coins and a certain fixed number N $\left( \ge 2 \right)$ of 10 and five paise coins.
Thus, total number of coins will be $2+5+N$
$=N+7$ ………………………(1)
Now, we are given the number of combinations of five coins whose total is less than one rupee and fifty paisa is $10\left( N+k \right)$ .
Let us consider event ‘E’ where the total sum of five coins is more than or equal to one rupee and fifty paisa. So, now we will do all the possible combinations of selection where the total amount is more than or equal to one rupee and fifty paisa.
First, taking one coin of fifty paise and 4 coins of twenty-five paisa $={}^{2}{{C}_{1}}\times {}^{5}{{C}_{4}}\times {}^{N}{{C}_{0}}$
Second, taking two coins of fifty paise and 3 coins of twenty-five paisa $={}^{2}{{C}_{2}}\times {}^{5}{{C}_{3}}\times {}^{N}{{C}_{0}}$
Third, taking two coins of fifty paisa, 2 coins of twenty-five paisa and one coin of 10 or five paisa $={}^{2}{{C}_{2}}\times {}^{5}{{C}_{2}}\times {}^{N}{{C}_{1}}$ .
So, total we get as
$={}^{2}{{C}_{1}}\times {}^{5}{{C}_{4}}\times {}^{N}{{C}_{0}}+{}^{2}{{C}_{2}}\times {}^{5}{{C}_{3}}\times {}^{N}{{C}_{0}}+{}^{2}{{C}_{2}}\times {}^{5}{{C}_{2}}\times {}^{N}{{C}_{1}}$
Here, we will use the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ . On solving, we get as
$\begin{aligned} & =\dfrac{2!}{\left( 2-1 \right)!1!}\times \dfrac{5!}{\left( 5-4 \right)!4!}\times 1+\dfrac{2!}{\left( 2-2 \right)!2!}\times \dfrac{5!}{\left( 5-3 \right)!3!}\times 1+ \\\ & \dfrac{2!}{\left( 2-2 \right)!2!}\times \dfrac{5!}{\left( 5-2 \right)!2!}\times \dfrac{N!}{\left( N-1 \right)!1!} \\\ \end{aligned}$
$\begin{aligned} & =\dfrac{2!}{1!1!}\times \dfrac{5\cdot 4!}{1!4!}\times 1+\dfrac{2!}{0!2!}\times \dfrac{5\cdot 4\cdot 3!}{2!3!}\times 1+ \\\ & \dfrac{2!}{0!2!}\times \dfrac{5\cdot 4\cdot 3!}{3!2!}\times \dfrac{N\cdot \left( N-1 \right)!}{\left( N-1 \right)!1!} \\\ \end{aligned}$
Further simplifying, we get as
$=2\times 5\times 1+1\times 10\times 1+1\times 10\times N$ (we should know that 0! is equal to 1)
$=10+10+10N$
$=10\left( N+2 \right)$
Thus, on comparing this with our equation given in question i.e. $10\left( N+k \right)$ , we can say that the value of k is 2.
Thus, the value of k is 2.

Note: Remember that we have to take complement of the statement total value of these 5 coins is less than one rupee and fifty paisa i.e. given as total value of 5 coins is more than or equal to one rupee and fifty paisa. If students find the value of the original statement given, then the answer will be e${}^{N+7}{{C}_{5}}-\left( N+2 \right)$ . We can also find the probability of event E by taking ${}^{N+7}{{C}_{5}}$ as the sample size. . By this also, we can compare only the numerator part and get the value of k as per asked in question.