A correspondent writes 7 letters and addresses 7 envelopes, one for each letter. In how many ways ca

7! * ( 1/2! - 1/3! + 1/4! - 1/5! + 1/6! - 1/7! )

= 2520 - 840 + 210 - 42 + 7 - 1

= 1854

ans is 18 54
no letter in right envelope=
n![1-1/1!+1/2!-1/3!+.......1/n!]
here n= 7

by using the formula D(n)=n! * (1/2! - 1/3! +1/4! - .... +(-1)^n/(n!) )

Hence D(7)=1854

1854
formula is n!(i/2!-1/3!+......+(-1)^n/n!)

n!(1/2!-1/3!+1/4!........(-1)^n*1/n!

here n=7

Number of ways in which 'n' objects can be placed on 'n' positions in such a manner that none of them is correct is given by the Dearrangement formula.

D(n)= n!(1/2!-1/3!+1/4!........(-1)^n*1/n!).

Dearr(7) = 7! * (1 - 1 + 1/2 - 1/6 + 1/24 - 1/120+1/6!-1/7!)

ans- 1854

According to this mathematical formula: 7! (1/2 ! - 1/3 ! + 1/4 ! - 1/5 ! + 1/6 ! - 1/7) = 1854

6 cases for the first, 5 for the second, 4 .. Wait a bit--suppose the first and second just swapped? Start from the beginning.

1 letter, always right, =0
2 letters, one way to goober it up
3 letters, 2 ways
4 letters: ah. 3*2=6 ways of cycling the 4 around, but then 3 ways of doing 2+2. (=9)

So partitioning is the way to go. A partition of 1 always maps the letter into the right envelope, so there's no answers with partition 1.

7 can be
a "cycle" of 7: with 1 case, which has 6*5*4*3*2 permutations
a "cycle" of 5 plus a "cycle" of 2: with 7*6/2 cases and 4*3*2 permutations
a "cycle" of 4 plus a "cycle" of 3: where I'm going to cheat and include the 2+2 in the 4, with 7*6*5/6 cases and (2 {for the 3} * 9 {for the 4}) permutations

That gives me 720+504+630=1854

for dearrangement Dn=n!(1/2!−1/3!+1/4!−....)
here n=7
Dn=7!(1/2!-1/3!+1/4!-1/5!+1/6!-1/7!)
=1852.2...
so nearest is 1854

According to the mathematical formula: 7!(1/2! -1/3! + 1/4!- 1/5! +1/6! -1/7!)=1854
option 7

from formula of dearrangement we get
n![1-1/1!+1/2!-1/3!+1/4!-1/5!+1/6!-1/7!]
n=7
we get ans as 1854..

if there are n things in a row, a permutation of these n things such that none of them occupies its original position. then total no of ways =
n!((1/2!)-(1/3!)+(1/4!)-(1/5!)+......+(-1)^n(1/n!))

D(N)=(N-1)[D(N-1)+D(N-2)]

AND D(2)=1,D(1)=0
ANS IS 1854

probability of placing atleast 1 letter into correct envelope + probability of placing none into correct envelope =1
so prob of placing atleast 1 into correct envelope is 1-1/2!+1/3!-1/4!+1/5!-1/6!+1/7! hence the other prob is 1-above result. i.e, 1/2!-1/3!+1/4!-1/5!+1/6!-1/7! =1854/5040.hence the num of ways =1854 out of 5040 ways.

7!*[ 1/2! - 1/3! + 1/4! - 1/5! + 1/6! - 1/7! ]
ans is 1854

7!(1-1/1!+1/2!-1/3!+1/4!-1/5!+1/6!-1/7!)
=1854

no. of de arrangements =n!{(1/2!)-(1/3!)...........(-1)^n/n!}

The number of derangements is: n!(1/2!-1/3!+.....+(-1)^n/n!)

as the formula is given by no of ways=n!(1-1/1!+1/2!-1/3!+....+1/n!)

on putting n=7 we get 1854.

according to formula
7!(1/2!-1/3!+1/4!-1/5!+1/6!-1/7!)
so on solving it deeply 1854 is the answer

7!{(1/2!)-(1/3!)+(1/4!)-(1/5!)+(1/6!)-(1/7!)}

to find the answer formula is n!(1-1/1!+1/2!-1/3!+.....(-1)1/n!)
so here n is 7
after solving this we get 1854
.

1854
According to this mathematical formula: 7! (1/2 ! - 1/3 ! + 1/4 ! - 1/5 ! + 1/6 ! - 1/7) = 1854

according to this mathematical formula:7!(1/2!-1/3!+1/4!-1/5!+1/6!-1/7!)=1854

according to permutataions and combinations
the output=7![(1/2)!-(1/3)!+(1/4)!-(1/5)!+(1/6)!-(1/7)!]=1854

According to this formula=7!*(1/2!-1/3!+1/4!-1/5!+1/6!-1/7!)

1 letter, always right, =0
2 letters, 1 way
3 letters, 2 ways
4 letters: ah. 3*2=6 ways of cycling the 4 around, but then 3 ways of doing 2+2. (=9)

So partitioning is the way to go. A partition of 1 always maps the letter into the right envelope, so there's no answers with partition 1.

7 can be
a "cycle" of 7: with 1 case, which has 6*5*4*3*2 permutations
a "cycle" of 5 plus a "cycle" of 2: with 7*6/2 cases and 4*3*2 permutations
a "cycle" of 4 plus a "cycle" of 3: where I'm going to cheat and include the 2+2 in the 4, with 7*6*5/6 cases and (2 {for the 3} * 9 {for the 4}) permutations

That gives Answer as 720+504+630=1854

Option 7)

6 cases for the first, 5 for the second, 4 .. Wait a bit--suppose the first and second just swapped? Start from the beginning.

1 letter, always right, =0
2 letters, one way to goober it up
3 letters, 2 ways
4 letters: ah. 3*2=6 ways of cycling the 4 around, but then 3 ways of doing 2+2. (=9)

So partitioning is the way to go. A partition of 1 always maps the letter into the right envelope, so there's no answers with partition 1.

7 can be
a "cycle" of 7: with 1 case, which has 6*5*4*3*2 permutations
a "cycle" of 5 plus a "cycle" of 2: with 7*6/2 cases and 4*3*2 permutations
a "cycle" of 4 plus a "cycle" of 3: where I'm going to cheat and include the 2+2 in the 4, with 7*6*5/6 cases and (2 {for the 3} * 9 {for the 4}) permutations

That gives me 720+504+630=1854

let take 2 envelopes .for above condition ,possibility is 1
for 3 possibility is 2
for 4 it is 8
.
.
.
.
.

for 7 it wil be1854

a/c to mathematical formula=7!(1/2!-1/3!+1/4!-1/5!+1/6!-1/7!)
=1854