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!

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
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
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
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

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