
The total variety of the way of choosing six coins out of twenty-one rupee coins, ten fifty paise coins, and seven twenty 5 paise coins is
A). 28
B). 56
C). \[{\;^{37}}{C_6}\]
D). None of these
Answer
219.3k+ views
Hint: First of all, find how many types of coins are available. And determine the numbers of coins that are to be selected out of the total coins. Therefore, the total number of selecting coins may be determined as \[{\;^{n + r - 1}}{C_r}\] Where n = types of the coins and r = number of coins has to be selected.
Complete step by step solution:
Let us consider that if there are r objects selected from a container and there are n types of objects available in the container, then the number of ways to select the object from the container may be determined as the formula which is given below.
\[\;{ \Rightarrow ^{n + r - 1}}{C_r}\]
In the question, we have given that
There are three types of the coins such as,
1 rupee coin, 50 paise coins and 5 paise coins. Therefore, the number of types of coins are,
\[\begin{array}{*{20}{c}}
{ \Rightarrow n}& = &3
\end{array}\]
And there are six coins selected out of 37 coins. Therefore,
\[\begin{array}{*{20}{c}}
{ \Rightarrow r}& = &6
\end{array}\]
Now, we know that the number of ways for selecting the objects from the container is,
\[{ \Rightarrow ^{n + r - 1}}{C_r}\]
And then we can write
\[{ \Rightarrow ^{3 + 6 - 1}}{C_6}\]
\[{ \Rightarrow ^8}{C_6}\]
Therefore we know that
\[\;\begin{array}{*{20}{c}}
{{ \Rightarrow ^n}{C_r}}& = &{\dfrac{{n!}}{{r!(n - r)!}}}
\end{array}\]
So,
\[\;\begin{array}{*{20}{c}}
{{ \Rightarrow ^8}{C_6}}& = &{\dfrac{{8!}}{{6!2!}}}
\end{array}\]
\[\;\begin{array}{*{20}{c}}
{{ \Rightarrow ^8}{C_6}}& = &{\dfrac{{8 \times 7 \times 6!}}{{6! \times 2 \times 1}}}
\end{array}\]
\[\;\begin{array}{*{20}{c}}
{{ \Rightarrow ^8}{C_6}}& = &{28}
\end{array}\]
Now, the final answer is 28.
Therefore, the correct option is (A).
Note: As per the given question, the coins are repeating which means repetition of the coins is allowed. That is why we will use the formula that is \[{\;^{n + r - 1}}{C_r}\].
Complete step by step solution:
Let us consider that if there are r objects selected from a container and there are n types of objects available in the container, then the number of ways to select the object from the container may be determined as the formula which is given below.
\[\;{ \Rightarrow ^{n + r - 1}}{C_r}\]
In the question, we have given that
There are three types of the coins such as,
1 rupee coin, 50 paise coins and 5 paise coins. Therefore, the number of types of coins are,
\[\begin{array}{*{20}{c}}
{ \Rightarrow n}& = &3
\end{array}\]
And there are six coins selected out of 37 coins. Therefore,
\[\begin{array}{*{20}{c}}
{ \Rightarrow r}& = &6
\end{array}\]
Now, we know that the number of ways for selecting the objects from the container is,
\[{ \Rightarrow ^{n + r - 1}}{C_r}\]
And then we can write
\[{ \Rightarrow ^{3 + 6 - 1}}{C_6}\]
\[{ \Rightarrow ^8}{C_6}\]
Therefore we know that
\[\;\begin{array}{*{20}{c}}
{{ \Rightarrow ^n}{C_r}}& = &{\dfrac{{n!}}{{r!(n - r)!}}}
\end{array}\]
So,
\[\;\begin{array}{*{20}{c}}
{{ \Rightarrow ^8}{C_6}}& = &{\dfrac{{8!}}{{6!2!}}}
\end{array}\]
\[\;\begin{array}{*{20}{c}}
{{ \Rightarrow ^8}{C_6}}& = &{\dfrac{{8 \times 7 \times 6!}}{{6! \times 2 \times 1}}}
\end{array}\]
\[\;\begin{array}{*{20}{c}}
{{ \Rightarrow ^8}{C_6}}& = &{28}
\end{array}\]
Now, the final answer is 28.
Therefore, the correct option is (A).
Note: As per the given question, the coins are repeating which means repetition of the coins is allowed. That is why we will use the formula that is \[{\;^{n + r - 1}}{C_r}\].
Recently Updated Pages
Chemical Equation - Important Concepts and Tips for JEE

JEE Main 2022 (July 29th Shift 1) Chemistry Question Paper with Answer Key

Conduction Explained: Definition, Examples & Science for Students

Analytical Method of Vector Addition Explained Simply

Atomic Size - Important Concepts and Tips for JEE

JEE Main 2022 (June 29th Shift 1) Maths Question Paper with Answer Key

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

Understanding Atomic Structure for Beginners

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections

NCERT Solutions for Class 11 Maths Chapter 9 Straight Lines

NCERT Solutions For Class 11 Maths Chapter 8 Sequences And Series

How to Convert a Galvanometer into an Ammeter or Voltmeter

NCERT Solutions For Class 11 Maths Chapter 12 Limits And Derivatives

