Combinatorics Intermediate Challenges-II

Using complementary counting..... $\large \ total \ choices = 4! = 24$

$\large \ choices \ for \ all \ correct = 1$

$\large \ choices \ for \ 3 \ correct = 0$

$\large \ choices \ for \ 2 \ correct = 6$ (you correctly allocate 2 letters in ${4 \choose 2 }$ ways, then you are left with 1 possibility to satisfy the condition of 2 correctly sent letters)

$\large \ choices \ for \ 1 \ correct = 8$ (correctly send a letter, then you are left with 2 choices to incorrectly send the next, then you are left with 1 possibility to satisfy the condition of 1 correctly sent letter. This is done for 4 letters, so we calculate 2 $\times$ 4)

$\therefore\ p(none \ correctly \ sent ) = \frac{9}{24} = \frac{3}{8} = 37.5 \ percent$

$No\quad of\quad ways\quad to\quad dis\quad arrange\quad n\quad items\\ A=\left( n! \right) \left[ \frac { 1 }{ 0! } -\frac { 1 }{ 1! } +\frac { 1 }{ 2! } -\frac { 1 }{ 3! } +\frac { 1 }{ 4! } .....\frac { { (-1) }^{ n } }{ n! } \right] \\ In\quad our\quad case\quad n=4\\ A=\left( 4! \right) \left[ \frac { 1 }{ 0! } -\frac { 1 }{ 1! } +\frac { 1 }{ 2! } -\frac { 1 }{ 3! } +\frac { 1 }{ 4! } \right] =9\\ Now\quad Probability\quad =\frac { 9 }{ 4! } =0.375\\ Percentage\quad =\quad 37.5\quad \%$

Let the four letters be $\{A, B, C, D\}$ .

$1^{st}$ letter $A$ can be delivered to $3$ houses.

$2^{nd}$ letter corresponding to the house where letter $A$ had been delivered could be delivered to any $3$ remaining houses.

The remaining letters can be delivered in only one way.

Hence total ways to deliver letter such that no letter is delivered correctly $= 3 \times 3$

Required probability $=\frac{9}{4!} \times 100 = \frac{900}{24} = 37.5\%$