The total number of ways in which six '+' and four '-' signs can be arranged in a line such that no two '-' signs occur together is
A) \[\dfrac{{71}}{{31}}\]
B) \[61 \times \dfrac{{71}}{{31}}\]
C) \[35\]
D) none of these
VerifiedHint:Here we will find the number of ways in which the ‘+’ signs can be arranged in a row and then we will find the number of ways in which the number of ‘-‘ signs can be arranged in remaining number of places in a row such that no two ‘-‘ signs are together.
Complete step-by-step answer:
We have to find the number of ways in which ‘+’ signs can be arranged in a row leaving a space for a ‘-‘ sign between two ‘+’ signs as two ‘-‘ signs can’t be placed together.
Six ‘+’ signs can be arranged at six places in a row in 1 way
Now we need to find the number of ways in which the ‘-‘ signs can be arranged in the remaining empty spaces hence,
Total number of empty spaces = 7
Number of girls = 4
Therefore, the number of ways in which 4 ‘-‘ signs can be arranged at 7 places is given by:
\[{}^7C_4\] ways
Now since we know that \[{}^nCr\] is given by:
\[{}^nC_r = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
Therefore,
\[{}^7C_4 = \dfrac{{7!}}{{4!\left( {7 - 4} \right)!}} \]
\[{}^7C_4 = \dfrac{{7!}}{{4! \times 3!}} \]
\[{}^7C_4 = \dfrac{{7 \times 6 \times 5 \times 4!}}{{4! \times 3 \times 2 \times 1}} \]
\[{}^7C_4 = \dfrac{{7 \times 6 \times 5}}{{3 \times 2 \times 1}} \]
\[{}^7C_4 = 7 \times 5 \]
\[{}^7C_4 = 35\]
Now, the total number of ways in which both ‘+’ signs and ‘-‘ signs can be arranged in a row
\[= 1 \times 35 = 35 \]
Hence option (C) is the correct option.
Note: The formula for permutation(arrangement) of r numbers out of n numbers is given by:
\[{}^nPr = \dfrac{{n!}}{{\left( {n - r} \right)!}}\]
Also, the value of \[0!\] is 1.
Also, the formula for combinations of r numbers out of n numbers is given by:
\[{}^nCr = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
Since the ‘+’ and ‘-‘ signs are not distinct and need to be arranged , therefore combination can be used and not combination.
