
$8$ 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
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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.
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