How many 5-digit bank security Personal Identification Numbers (PIN) codes can be formed using the characters in the set {0, 2, 4, 6, 8, a, b, c, d, y } with the restrictions below.
Note: For purposes of this problem, the PIN number can be "padded" with zeros on the left. So a 0z01a is a good PIN. (IE, PIN codes can start with zeros.)
FYI: The 5-position pin above would rate as "weak" in most systems. Most of them require at least 8 characters from the digits, upper and lower case letters and special characters, and usually do not want a reused character!
Each part below is independent of the previous parts.
(a) PINs must use exactly three a's among the five characters. (with replacement)
(b) PINs begin with the letters a or d and can repeat characters. (with replacement)
(a) To form a PIN with exactly three 'a's among the five characters, we need to consider the possible positions for the 'a' characters. Since there are 5 positions and we need to choose 3 for 'a', this can be done in 5 choose 3 ways, which is equal to (5! / (3!(5-3)!)) = 10 ways.
Now, for the remaining 2 characters, we can choose any character from the set {0, 2, 4, 6, 8, b, c, d, y }. Since each character can be chosen independently and there are 9 options, we have 9 choices for each character. Therefore, the total number of PINs in this case is 10 * 9 * 9 = 810.
(b) To form a PIN that begins with either 'a' or 'd' and can have repeated characters, we have the following possibilities for the first character: 'a' or 'd'. For the remaining 4 characters, we can choose any character from the set {0, 2, 4, 6, 8, a, b, c, d, y }. Since each character can be chosen independently and there are 10 options, we have 10 choices for each character.
The total number of PINs in this case is 2 * 10 * 10 * 10 * 10 = 20,000.
So, there are 810 PINs that meet the restrictions in part (a) and 20,000 PINs that meet the restrictions in part (b).
To find the answer to each part, we can use the concept of permutations. In a permutation, the order of arrangement matters and repetition is allowed.
(a) PINs must use exactly three a's among the five characters.
In this case, we have to arrange 3 a's and 2 characters from the set {0, 2, 4, 6, 8, b, c, d, y}. The total number of possible PINs can be found using the formula for permutations:
nPr = n! / (n - r)!
Here, n is the total number of available characters, which is 9 (since we have 5 digits and 4 non-digit characters), and r is the number of characters to be arranged, which is 5.
Applying the formula, we get:
9P5 = 9! / (9 - 5)!
= 9! / 4!
= (9 * 8 * 7 * 6 * 5!) / 4!
= 9 * 8 * 7 * 6
= 3,024
Therefore, there are 3,024 5-digit bank security PIN codes that can be formed using the characters in the set {0, 2, 4, 6, 8, a, b, c, d, y} with exactly three a's.
(b) PINs begin with the letters a or d and can repeat characters.
In this case, we have two possibilities for the first character: a or d. For each of these possibilities, we can choose any character from the set {0, 2, 4, 6, 8, a, b, c, d, y} to fill the remaining 4 positions.
The total number of possible PINs can be found by summing the number of PINs for each possibility:
Number of PINs starting with 'a':
1 * 10^4 = 10,000
Number of PINs starting with 'd':
1 * 10^4 = 10,000
Total number of possible PINs = Number of PINs starting with 'a' + Number of PINs starting with 'd' = 10,000 + 10,000 = 20,000
Therefore, there are 20,000 5-digit bank security PIN codes that can be formed using the characters in the set {0, 2, 4, 6, 8, a, b, c, d, y} with PINs that begin with the letters a or d and can repeat characters.
To answer these questions, we need to apply the principles of permutations and combinations.
For part (a), we are looking for the number of 5-digit PIN codes that have exactly three 'a' characters among the five characters. Since we have three 'a's, we can choose the positions they will occupy in the PIN code in (5 choose 3) ways. Once we have decided the positions of the 'a's, we need to fill the remaining two positions with any of the remaining characters from the given set.
The formula for combinations is nCr = n! / (r! * (n-r)!), where n is the total number of elements and r is the number of elements we want to choose.
In this case, n = 5 (total number of positions) and r = 3 (number of 'a's). Thus, (5 choose 3) gives us 5! / (3! * (5-3)!) = 10.
Now, each of the remaining two positions can be filled with any of the remaining 9 characters (excluding the three 'a's that have already been placed). Therefore, the total number of 5-digit PIN codes that satisfy this condition is 10 * 9 * 9 = 810.
For part (b), we need to find the number of 5-digit PIN codes that begin with the letters 'a' or 'd' and allow repeated characters.
To calculate this, we need to consider two cases:
1) PIN codes that begin with 'a'
2) PIN codes that begin with 'd'
For both cases, we have one fixed character at the beginning of the PIN code. For the remaining four positions, we can choose any character from the given set, including repetitions. Therefore, we have 10 choices for each position.
In the case of PIN codes that begin with 'a', we have one fixed character and four remaining positions that can be filled with any of the 10 characters. Thus, there are a total of 10^4 = 10,000 PIN codes that satisfy this condition.
Similarly, in the case of PIN codes that begin with 'd', there are also 10^4 = 10,000 PIN codes that satisfy this condition.
Therefore, the total number of 5-digit PIN codes that begin with either 'a' or 'd' and can have repeated characters is 10,000 + 10,000 = 20,000.