Gap and string method in permutations and combinations Can somebody please explain the gap and the string method used in permutations and combinations (combinatorics) to solve certain types of problems such as finding the number of arrangements when a number of objects are not together or a number of objects are always together. My teacher was talking about these and I could not understand what he was saying. I searched the internet but could not find good resources.
More information: I am in grade 11 currently and this question is relevant to the chapter Permutations and Combinations which we have in our Mathematics course. More specifically this question is relevant to questions like finding the number of arrangements when out of all objects at hand some particular objects should not be adjacent and the number of arrangements when out of all objects at hand some particular objects should always occur together. I have a basic understanding of Permutations and Combinations.
 A: Take the word SIGNATURE, and find permutations if
(a): no two vowels are together (used gap method)
(b): The letters S,G,N are together (used string method)
(a) There are $4$ vowels which can be placed in the gaps between consonants in $\binom64$ ways, and the vowels and consonants can be permuted separately.
$- S - G - N - T - R - \;\; \binom64 \times 4! \times 5!$
(b) Clump $SGN$ into a string $\boxed {SGN}$ so there are $1 + 6 =7$ "objects"
The clumped string can be permuted in $3!$ ways, and the $7$ objects in $7!$ ways,  thus $7!*3!$ ways
There can be many variations of such problems, and all letters may not be distinct as they are here, but the basic principles of the two methods will remain the same.
A: These two methods are commonly used method.we will explain it through an example.
Suppose you have the word "SCHOOL" 
(i) find total possible arrangement if both O are never together(Gap method) 
(ii) CHL always come together(string or tie method) 
Now,coming to the first question we have to use the gap method. Now there are four more letters after removing O now we get 3 gaps between these letters,so we can keep the two O in (3,2) ways 3C2 ways..  now permuting remaining letters ..we get 4! And total number of possible arrangement will be 3C2*4! .. here one crucial thing to know...we don't have to permute OO since both are same and their permute will be 1 
 S__C__H__L 
Now coming to the second question.. Here CHL should be 
Together..so we will tie the CHL with a string ,so that total number of objects are 6 .. now permuting these 6 objects we get 6!/2! ,but intra permutation is also possible between CHL hence 3! .now total number of ways = 6!*3!/2! .. here numerator is divided  by 2! because OO are the same thing.... 
So yeah... hope it's clear to you... 
