
$8$ identical coins are arranged in a row. The total number of ways in which the number of heads is equal to the number of tails is?
Answer
164.4k+ views
Hint: For simplifying the problem based on permutation and combination, we know that ‘r’ items can be selected from a collection of ‘n’ items in $^n{C_r}$ ways. According to the question first, we will find the number of total possible ways for the number of heads and tails to be equal to get the solution to the given problem.
Formula used:
The formula used in this problem to find out the number of ways from permutation is $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$.
Complete Step by Step Solution:
It is given that the number of heads is equal to the number of Tails.
$Number{\text{ }}of{\text{ }}heads = Number{\text{ }}of{\text{ }}Tails$
Let H represent the number of heads and T represent the number of tails.
So, the required number of ways must be equal to the number of ways in which letters H and T can be arranged like this HHHHTTTT (the positions can be different).
Also, we know the formula to find out the number of ways from permutation is $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Now, we know that there is total $8$ coins $(n)$ . But for the same number of heads and tails, half of them i.e., $4$ must be heads and 4 must be tails $(r)$ .
Thus, the total number of ways is ${}^8{C_4} = \dfrac{{8!}}{{4!\left( {8 - 4} \right)!}} = \dfrac{{8!}}{{4! \times 4!}}$
Now, Open the factorial by using the formula $n! = n(n - 1)(n - 2)(n - 3)............$
$ = \dfrac{{8 \times 7 \times 6 \times 5 \times 4!}}{{4 \times 3 \times 2 \times 1 \times 4!}} = 70$
Hence, the total number of ways in which the number of heads equals the number of tails is $70$ .
Note: In this problem, it is necessary to analyze the given conditions on the basis of which the number of total ways of different combinations is calculated. Also, it is advised to avoid considering the other ways where $(Number{\text{ }}of{\text{ }}heads \ne Number{\text{ }}of{\text{ }}Tails)$ in order to perform the calculation part precisely, otherwise we will get an incorrect answer.
Formula used:
The formula used in this problem to find out the number of ways from permutation is $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$.
Complete Step by Step Solution:
It is given that the number of heads is equal to the number of Tails.
$Number{\text{ }}of{\text{ }}heads = Number{\text{ }}of{\text{ }}Tails$
Let H represent the number of heads and T represent the number of tails.
So, the required number of ways must be equal to the number of ways in which letters H and T can be arranged like this HHHHTTTT (the positions can be different).
Also, we know the formula to find out the number of ways from permutation is $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Now, we know that there is total $8$ coins $(n)$ . But for the same number of heads and tails, half of them i.e., $4$ must be heads and 4 must be tails $(r)$ .
Thus, the total number of ways is ${}^8{C_4} = \dfrac{{8!}}{{4!\left( {8 - 4} \right)!}} = \dfrac{{8!}}{{4! \times 4!}}$
Now, Open the factorial by using the formula $n! = n(n - 1)(n - 2)(n - 3)............$
$ = \dfrac{{8 \times 7 \times 6 \times 5 \times 4!}}{{4 \times 3 \times 2 \times 1 \times 4!}} = 70$
Hence, the total number of ways in which the number of heads equals the number of tails is $70$ .
Note: In this problem, it is necessary to analyze the given conditions on the basis of which the number of total ways of different combinations is calculated. Also, it is advised to avoid considering the other ways where $(Number{\text{ }}of{\text{ }}heads \ne Number{\text{ }}of{\text{ }}Tails)$ in order to perform the calculation part precisely, otherwise we will get an incorrect answer.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

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 4 Complex Numbers and Quadratic Equations

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets

Instantaneous Velocity - Formula based Examples for JEE

NCERT Solutions for Class 11 Maths Chapter 8 Sequences and Series
