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Question: The number of ways in which p positive signs may be placed in a row so that no two negative signs sh...

The number of ways in which p positive signs may be placed in a row so that no two negative signs shall be together is
(A) p+1Cn{}^{p + 1}{C_n}
(B) pCn{}^p{C_n}
(C) p−1Cn{}^{p - 1}{C_n}
(D) None of these

Explanation

Solution

Start with placing ‘p’ plus signs in a row with a gap of one in between two ′+′' + ' signs. This will give us ′p+1′'p + 1' places to arrange the ′−′' - ' sign in them without keeping two of them together. Now use the concept of combinations to arrange ‘n’ identical negative signs on ′p+1′'p + 1' places. Match the calculated answer with the options to find the correct one.

Complete step-by-step answer:
Here in this problem, we need to find the number of ways in which ‘p’ positive signs (+)\left( + \right) and ‘n’ negative signs (−)\left( - \right) can be arranged in a row so that no two negative signs are together. And along with this information we are given with four options to choose as the correct answer based on our calculations.
Since all the given options are in the form of combinations (nCr)\left( {{}^n{C_r}} \right) , let us understand the concept of combinations before starting the solution. A combination is a mathematical technique that determines the number of possible arrangements in a collection of items where the order of the selection does not matter. In combinations, you can select the items in any order.
⇒\RightarrowThe combinations for arranging ‘r’ things on ‘n’ places are given by nCr{}^n{C_r} , i.e. nCr=n!r!(n−r)!{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}
In this case, we have to first arrange pp positive signs which are identical. This should be done leaving a gap of one in between each of the ′+′' + ' signs. This can be done in the only 11 way.
Therefore, we got p+1p + 1 gaps for arranging the ′−′' - ' signs without keeping two of them together.
To arrange the ′−′' - ' signs in the remaining gaps we need to arrange ‘n’ signs in p+1p + 1 places to satisfy the given condition
⇒\Rightarrow The combinations for arranging ‘n’ identical things on p+1p + 1 places can be written as p+1Cn{}^{p + 1}{C_n}
Hence, the option (A) is the correct answer.

Note: In questions like this, it is very important to analyse the details properly. It is important to notice that both signs here are identical and this allowed us to use combinations because here the order of the signs does not matter. If the signs were to be arranged while keeping the order in mind, we can take the use of permutations.