1. ## distinct permutations.............

hello,
plz try to do these questions.

(a) Find the number of distinct permutations that can be formed using the letters of the word “Mathematics”.

(b) How many integers from 1 through 1000 are multiples of 3 or multiples of 5?

Question 2:

Compute for each of the following values of x.
( a ) 20/3 ( b) -2.07

Question 3:

Two cards are drawn at random from a well shuffled pack of 52 cards. Find the probability that:
(i) one is king and other is queen.
(ii) Both are of same color.
(iii) Both are of different color.
Question 4:

A coin is biased such that a head is thrice as likely to occur as a tail. Find the probability distribution of heads and also find the mean and variance of the distribution when it is tossed 4 times.

2. For MATHEMATICS:

There are 11 letters. The M, A, and T are repeated twice.

$\frac{11!}{2!2!2!}=6,652,800$

#2:

There are 333 multiples of 3 from 1 to 1000

There are 200 multiples of 5 from 1 to 1000

There are 66 which are a mutiple of both.

200+333-66=467

3. Hello, m777!

Here's #3 . . .

Two cards are drawn at random from a well-shuffled pack of 52 cards.
Find the probability that:
. . (a) one is king and other is queen.
. . (b) Both are of same color.
. . (c) Both are of different color.

(a) one King and one Queen
The first card must be a King or a Queen ... probability $\frac{8}{52}$
The second card must one of the other value ... probability $\frac{4}{51}$

Hence: . $P(K \land Q) \:=\:\frac{8}{52}\cdot\frac{4}{51}\:=\:\boxed{\fra c{8}{663}}$

(b) Same color
The first card can be any card ... probability $\frac{52}{52} = 1$
The second card must be one of the same color ... probability $\frac{25}{51}$
Hence:. $P(\text{same color}) \:= \:1\cdot\frac{25}{51}\:=\:\boxed{\frac{25}{51}}$

(c) Different colors
$P(\text{different colors}) \;=\;1 - P(\text{same color}) \;=\;1 - \frac{25}{51} \;=\;\boxed{\frac{26}{51}}$

4. Originally Posted by m777

(b) How many integers from 1 through 1000 are multiples of 3 or multiples of 5?
Inclusion-Exclusion principle is your friend here.

Let $S$ be the set of all numbers 1 through 1000 that are either 3 or 5 multiples.

Let $S_3$ be only 3 multiples.

Let $S_5$ be only 5 multiples.

$S=S_3\cup S_5$.

Thus, (Inclusion-Exclusion),
$|S_3\cup S_5|=|S_3|+|S_5|-|S_3\cap S_5|$.
Note,
$|S_3|=[1000/3]=333$
$S_5=[1000/5]=200$
And,
$S_3\cap S_5=S_{15}$
because $\gcd(3,5)=1$.
Note,
$S_{15}=[1000/15]=66$
$333+200-66$