Question

The number of ways of arranging m positives and n (< m+l) negative signs in a row so that no two negative signs are together is

& A.\text{ }{}^{m+1}{{P}_{n}} \\\ & B.\text{ }{}^{n+1}{{P}_{m}} \\\ & C.\text{ }{}^{m+1}{{C}_{n}} \\\ & D.\text{ }{}^{n+1}{{C}_{m}} \\\ \end{aligned}$$

Answer 1

Arrange m positive signs, as there are no restrictions you can place it anywhere in the row. Now find the number of gaps that forms while arranging positive signs. Hence, find a number of ways to put n signs in the gap using combination, as order doesn’t matter here.

Complete step by step answer:
Let us first arrange the positive signs. We have not been given any kind of restrictions concerning the arrangements of positive signs. Thus, we can arrange positive signs in whatever place we want.
Now we have to arrange the negative signs in a way that no two signs are together. We can arrange it place a negative sign between two positive sign as shown below,
$+-+-+-+\ldots \ldots ..$ We can also put it as
$-+++-+-++-\ldots \ldots$
We can form a lot of expressions like this.
Now, if there are m positive signs, then there are (m+1) gaps for example, there are 3 positive signs, thus the gaps in between them are given as,
$-+-+-+-$
Hence, for 3 positive signs, there are 4 gaps. Similarly for m positive signs there are (m+1) gaps. Now, out of these (m+1) gaps are needed to put in negative signs. Thus we can put n signs in (m+1) gaps as a form of combination, as the need not be ordered.
Combination is of the form ${}^{n}{{C}_{r}}$.Thus we can put n signs in (m+1) gaps in ${}^{m+1}{{C}_{n}}$ways.
Therefore, a number of ways of arranging m positive and negative signs are in ${}^{m+1}{{C}_{n}}$ ways.

So, the correct answer is “Option C”.

Note:
All the negative signs are identical. Hence, it doesn’t matter where we put the n negative signs in the (m+1) gaps. Thus the order or arrangement of the sign doesn’t matter here. So we use combinations for this particular question.