### Permutations under Restrictions

**Case 1: When s particular things are always to be included**Number of permutations of n different 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 gives the formula as^{(n-s)}P_{(r-s)}×^{r}P_{s}which is same as^{(n-s)}C_{(r-s)}× r!

^{(n-s)}C_{(r-s)}× r! = $MF#%\dfrac{(n-s)! \times r!}{[n-s-(r-s)]! \times (r-s)!} = \dfrac{(n-s)! \times r!}{(n-r)! \times (r-s)!} $MF#%

^{(n-s)}P_{(r-s)}×^{r}P_{s}= $MF#% \dfrac{(n-s)!}{[n-s-(r-s)]!} ] \times \dfrac{r!}{(r-s)!} = \dfrac{(n-s)! \times r!}{(n-r)! \times (r-s)!}$MF#%

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

**Case 2: When a particular thing is always to be included**Number of permutations of n different 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 different things taking r at a time, when s particular things are never included is

^{(n-s)}C_{r}× 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)}C_{r}ways.

r objects can be arranged in r! ways

Total number of arrangements =^{(n-s)}C_{r}× r!**Case 4: When a particular thing is never included**Number of permutations of n different things taking r at a time, when a particular thing is never included, is

^{(n-1)}C_{r}× 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 different 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 different 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 different objects taking all at a time = n!

Number of arrangements of n different 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 different things taking all at a time, when m particular things never come together = n! - (n-m+1)! × m!

Comments(4)

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

Regards,

Support Team, careerbless.com

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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