Question

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

Answer 1

Hint: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.