Nine cards are numbered 1, 2, 2, 3, 3, 4, 6, 6, 6.Three of the nine cards are chosen and placed in a line, making a 3-digit number. Find how

many different numbers can be made in this way if the number is between 200 and 300.

Tell me answer in terms of permutation and combinations.

The first digit is 2

That leaves 5 unique digits (12346) to choose from, giving 5P2 = 20
But you have to add the 2 numbers with repeated digits: 233 and 266
total: 22 choices

We have 5 digits that are unique in which we have to chose from. We need to get two digits from five to get (5P2) numbers to be formed. We have two digits that are repeated in the range that is 6 and 3. and so we must add 2P2. in total we will have a total of 5P2+2P2 numbers formed. This sums to 22 numbers.

To solve this problem, we can use the principles of permutations and combinations.

First, let's determine the number of ways three cards can be chosen out of the nine available cards. This can be calculated using combinations. Since we are choosing three cards from nine without regard to their order, we can use the combination formula:

C(n, r) = n! / (r!(n - r)!)

In this case, n = 9 (number of available cards) and r = 3 (number of cards to be chosen). So, the number of ways to choose three cards is:

C(9, 3) = 9! / (3!(9 - 3)!) = 84

Now, let's calculate the number of possible three-digit numbers that can be formed using these three chosen cards.

Since the order matters, we can use permutations. We have three digits to be arranged, so we can use the permutation formula:

P(n, r) = n!

In this case, n = 3 (number of digits) and r = 3 (number of digits to be arranged). So, the number of possible arrangements of three digits is:

P(3, 3) = 3! = 6

However, we need to consider that the chosen number must be between 200 and 300.

Since the hundreds digit can only be 2, we have fixed one digit. Now we have two digits left to arrange in the remaining two spots (tens and units places).

So, the number of possible arrangements without considering the restrictions is:

P(2, 2) = 2! = 2

To find the total number of different numbers that can be made within the given range, we multiply the number of ways to choose three cards (combinations) by the number of possible arrangements (permutations):

Total = C(9, 3) * P(2, 2) = 84 * 2 = 168

Therefore, there are 168 different numbers that can be made if the number is between 200 and 300 using the given set of nine cards.

Well, if we want the number to be between 200 and 300, the hundreds place has to be 2. So, now we need to choose two more cards from the remaining eight cards.

The number of ways to choose two cards from eight is given by the combination formula: C(n, r) = n! / (r! * (n-r)!)

Plugging in the values, we get C(8, 2) = 8! / (2! * (8-2)!) = 28.

Now, for each combination of two cards, we can arrange them in 2! = 2 ways, as the two remaining cards have identical numbers.

Therefore, the total number of different numbers that can be made is 28 * 2 = 56.

So, there are 56 different numbers that can be made if the number is between 200 and 300.

To solve this problem using permutations and combinations, we can break it down into steps:

Step 1: Determine the total number of ways to choose 3 cards from the given set of 9 cards. This can be calculated using combinations.

Step 2: Determine the number of ways to arrange the chosen 3 cards to form a 3-digit number.

Step 3: Count the number of 3-digit numbers that fall within the given range (200 to 300).

Let's go through each step:

Step 1: Determining the combinations

Given that there are 9 cards and we need to choose 3 from them, we can calculate the number of combinations using the formula:
C(n, r) = n! / (r!(n-r)!)

In this case, n = 9 (number of cards) and r = 3 (cards chosen).

C(9, 3) = 9! / (3!(9-3)!) = 9! / (3!6!) = (9 * 8 * 7) / (3 * 2 * 1) = 84

So, there are 84 different ways to choose 3 cards from the given set.

Step 2: Determining the arrangements

Since each of the 3 chosen cards represents a digit in the 3-digit number, the number of arrangements can be calculated using permutations.

P(n, r) = n!

In this case, n = 3 (since we have 3 cards chosen).

P(3, 3) = 3! = 3 * 2 * 1 = 6

So, there are 6 different ways to arrange the chosen 3 cards.

Step 3: Counting the numbers within the given range

To calculate the number of 3-digit numbers between 200 and 300, we need to consider the restrictions:

- The hundreds digit can only be 2.
- The tens digit can be any of the chosen 3 cards (since they are repeated).
- The units digit can be any of the chosen 3 cards (since they are repeated).

Since there are no restrictions on the arrangement of the tens and units digits (as they are repeated cards), we can simply multiply the number of arrangements (6) by the number of choices for the tens (3) and the number of choices for the units (3).

Total number of 3-digit numbers = 6 * 3 * 3 = 54

So, there are 54 different numbers that can be made in this way, considering the given range, using permutations and combinations.