How many positive, three-digit integers contain at least one $3$ as a digit but do not contain a $5$ as a digit?

One such number is 333.

The digits to be considered are 0,1,2,4,6,7,8,9 as well as 3.

Numbers containing 2 3's :- 3.3 - there are 8 of these; 33. - there are 8 and .33 - there are 7 ( because the first number can't be 0)

Numbers beginning with a 3 + 2 other digits: 3.. There are 8P2 + 8 ( these last 8 are the double digits)

Numbers .3. There are 7P2 + 7 and Numbers ..3 there are 7P2 + 7 also.

7P2 = 7! / 5! = 42, 8P2 = 8! / 6! 56

So the answer is 1 + 8 + 8 + 7 + 56 + 8 + 42 + 7 + 42 + 7

= 186

200

Step-by-step explanation:

Let us consider the number of three-digit integers that do not contain $3$ and $5$ as digits; let this set be $S$. For any such number, there would be $7$ possible choices for the hundreds digit (excluding $0,3$, and $5$), and $8$ possible choices for each of the tens and ones digits. Thus, there are $7 \cdot 8 \cdot 8 = 448$ three-digit integers without a $3$ or $5$.

Now, we count the number of three-digit integers that just do not contain a $5$ as a digit; let this set be $T$. There would be $8$ possible choices for the hundreds digit, and $9$ for each of the others, giving $8 \cdot 9 \cdot 9 = 648$. By the complementary principle, the set of three-digit integers with at least one $3$ and no $5$s is the number of integers in $T$ but not $S$. There are $648 - 448 = \boxed{200}$ such numbers.

answer:

step-by-step explanation:

here, the total dishes = 10

dishes having meat = 3 ( chicken, meatloaf and hamburger )

thus, the probability of having meat on the chosen dishes = number of dishes having meat / total dishes

=

therefore, the probability of not having meat = 1- the probability of having meat on the chosen dishes

=

=

=

⇒ option b is correct.