Prove this result:?

Question

The number of ways in which "p" positive and "n" negative signs may be placed in a row so that no two negative signs shall be together is:

(p+1) C (n)

This subject involves permutations and combinations. Please help me out here. Thank you very much in advance.

Answer 1

Assume that the 'p' +ve signs r placed in alternate positions with blanks like this
_+_+_+_+...........................................+_+_
Now there r 'p' positive signs and (p+1) blanks.
(p+1) is the maximum no of blanks that can be put inside +ve signs.So the no of ways of selecting 'n' blanks from p+1 blanks is (p+1)C n.
Ex)If there are 3 positive signs & 1 negative sign,then the -ve signs can come in the following situations
+-++
-+++
++-+
+++-
So the no of ways by which the blanks can come is (p+1=3+1=4).So the answer is 4 C1=4.

Answer 2

An interesting, although obscure question. You will have a diificult time finding one person out of 6 billion who would even care about it....I have asked questions like this.

Answer 3

first arrange p + signs among themseleves in p!/p!=1way
now there are( p+1)places for n - signs they can be arranged in (p+1)p(n)/n!=(p+1)c(n)*n!/n!=(p+1)(c)n