Permutations under RestrictionsCase 1: When s particular things are always to be included

Number of permutations of n distinct things taking r at a time, when s particular things are always to be included in each arrangement, is
(n-s)C(r-s) × r!

Derivation of the formula
(r−s) objects can be selected from the (n−s) objects in (n-s)C(r-s) ways.

s objects can be selected from s objects only 1 way.

Totally r objects are selected and these can be arranged in r! ways.

Total number of arrangements = (n-s)C(r-s) × r!

Alternative Form
Some text books give the formula as (n-s)P(r-s) × rPs which is same as (n-s)C(r-s) × r!

(n-s)C(r-s) × r! @@=\dfrac{(n-s)! \times r!}{[n-s-(r-s)]! \times (r-s)!}@@ @@=\dfrac{(n-s)! \times r!}{(n-r)! \times (r-s)!} @@

(n-s)P(r-s) × rPs @@=\dfrac{(n-s)!}{[n-s-(r-s)]!} \times \dfrac{r!}{(r-s)!}@@ @@=\dfrac{(n-s)! \times r!}{(n-r)! \times (r-s)!}@@

As you can see, both these formulas are indeed the same.

Case 2: When a particular thing is always to be included

Number of permutations of n distinct things taking r at a time, when a particular thing is always to be included in each arrangement, is
(n-1)C(r-1) × r!

Derivation of the formula is similar to that of case 1.

Alternative Form
As we have seen for case 1, the same formula can also be expressed as (n-1)P(r-1) × r

Case 3: When s particular things are never included

Number of permutations of n distinct things taking r at a time, when s particular things are never included is
(n-s)Cr × r!

Derivation of the formula
Remove the s objects which are never included and we are left with (n-s) objects. r objects can be selected from these (n−s) objects in (n-s)Cr ways.

r objects can be arranged in r! ways.

Total number of arrangements
= (n-s)Cr × r!

Case 4: When a particular thing is never included

Number of permutations of n distinct things taking r at a time, when a particular thing is never included, is
(n-1)Cr × r!

Derivation of the formula is similar to that of case 3.

Case 5: When m particular things always come together

Number of permutations of n distinct things taking them all at a time, when m particular things always come together, is
(n-m+1)! × m!

Derivation of the formula
Group these m objects and consider it as a single object. Then the total number objects = (n-m+1). These (n-m+1) objects can be arranged in (n-m+1)! ways

m objects can be arranged in m! ways.

Total number of arrangements = (n-m+1)! × m!

Case 6: When m particular things never come together

Number of permutations of n distinct things taking them all at a time, when m particular things never come together, is
n! - (n – m + 1)! × m!

Derivation of the formula
Total number of arrangements possible using n distinct objects taking all at a time = n!

Number of arrangements of n distinct things taking all at a time, when m particular things always come together, is
(n-m+1)! × m!      (As we have seen in case 5)

Hence, number of permutations of n distinct things taking all at a time, when m particular things never come together
= n! - (n-m+1)! × m!

 
 
 
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showing 1-4 of 4 comments,   sorted newest to the oldest
vrish
2015-03-15 18:35:11 
In case 1 on this page - number of permutations of n different things taken r at a time when s things are to be included, - why should it be multiplied by (r-s+1)? In other words, what does the term imply. Can you please explain or derive the formula?

Thanks
(0) (0) Reply
careerbless.com
2015-03-19 21:49:27 
Dear Vrish,

Thanks for pointing this out. It was a typo error which is corrected with the derivation.

Regards,
Support Team, careerbless.com
(0) (0) Reply
jawed pathan
2015-03-14 05:33:24 
Sir please send me solution of this problem from permutation.

Q. In a meeting,everyone had shaken hands with everyone else. It was found that 66 hand shakes were exchange. How many numbers were present at the meeting?
Ans. 12
(0) (0) Reply
Jay
2015-03-14 09:23:22 
Assume that there are n persons

Number of hand shakes the first person can do with all others = (n-1)
Number of hand shakes the second person can do with all others = (n-2) (reducing one from (n-1) as handshake between first and second person is already counted)
this goes on

Total number of handshakes = (n-1)+(n-2)+...1 = n(n-1)/2
You can remember this formula. Coming back to the question

n(n-1)/2 = 66
n^2 - n - 132 = 0
(n-12)(n+11)=0
n=12

Required number of persons = 12
(0) (0) Reply
 
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