How many three digit numbers can be formed from the digits 2,4,5, 6, 7 if no digit is repeated

Problem 1: How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that – 

(i) repetition of the digits is allowed?

Solution: 

Answer: 125.

Method:

Here, Total number of digits = 5

Let 3-digit number be XYZ.

Now the number of digits available for X = 5,

As repetition is allowed,

So the number of digits available for Y and Z will also be 5 (each).

Thus, The total number of 3-digit numbers that can be formed = 5×5×5 = 125.

(ii) repetition of the digits is not allowed? 

Solution:

Answer: 60.

Method:

Here, Total number of digits = 5

Let 3-digit number be XYZ.

Now the number of digits available for X = 5,

As repetition is not allowed,

So the number of digits available for Y = 4 (As one digit has already been chosen at X),

Similarly, the number of digits available for Z = 3.

Thus, The total number of 3-digit numbers that can be formed = 5×4×3 = 60.

Problem 2: How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?

Solution:

Answer: 108.

Method:

Here, Total number of digits = 6

Let 3-digit number be XYZ.

Now, as the number should be even so the digits at unit place must be even, so number of digits available for Z = 3 (As 2,4,6 are even digits here),

As the repetition is allowed,

So the number of digits available for X = 6,

Similarly, the number of digits available for Y = 6.

Thus, The total number of 3-digit even numbers that can be formed = 6×6×3 = 108.

Problem 3: How many 4-letter code can be formed using the first 10 letters of the English alphabet if no letter can be repeated? 

Solution:

Answer: 5040

Method:

Here, Total number of letters = 10

Let the 4-letter code be 1234.

Now, the number of letters available for 1st place = 10,

As repetition is not allowed,

So the number of letters possible at 2nd place = 9 (As one letter has already been chosen at 1st place),

Similarly, the number of letters available for 3rd place = 8,

and the number of letters available for 4th place = 7.

Thus, The total number of 4-letter code that can be formed = 10×9×8×7 = 5040.

Problem 4: How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?

Solution:

Answer: 336

Method:

Here, Total number of digits = 10 (from 0 to 9)

Let 5-digit number be ABCDE.

Now, As the number should start from  67 so the number of possible digits at A and B = 1 (each),

As repetition is not allowed,

So the number of digits available for C = 8 ( As 2 digits have already been chosen at A and B),

Similarly, the number of digits available for D = 7,

and the number of digits available for E = 6.

Thus, The total number of 5-digit telephone numbers that can be formed = 1×1×8×7×6 = 336.

Problem 5: A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?

Solution:

Answer: 8

Method:

We know that, the possible outcome after tossing a coin is either head or tail (2 outcomes),

Here, a coin is tossed 3 times and outcomes are recorded after each toss,

Thus, the total number of outcomes = 2×2×2 = 8.

Problem 6: Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?

Solution:

Answer: 20.

Method:

Here, Total number of flags = 5

As each signal requires 2 flag and signals should be different so repetition will not be allowed,

So, the number of flags possible for the upper place = 5,

and the number of flags possible for the lower place = 4.

Thus, the total number of different signals that can be generated = 5×4 = 20.

Algebra ->  Permutations -> SOLUTION: a) How many three digit numbers can be formed from the digits 2,3,4,5,6,7 and 8? No digit repeats in the same number. b) What is the probability of an even number? c) What is the      Log On

Click here to see ALL problems on Permutations Question 1143267: a) How many three digit numbers can be formed from the digits 2,3,4,5,6,7 and 8? No digit repeats in the same number. b) What is the probability of an even number? c) What is the probability of numbers greater than 550 but less than 770? Found 2 solutions by math_helper, Edwin McCravy: Answer by math_helper(2357)     (Show Source): You can put this solution on YOUR website! There are 7 digits from which 3 must be selected, and order matters: (a) P(7, 3) = 7!/(7-3)! = 7!/4! = 7*6*5 = 210 You are only supposed to post one problem per-post, but I'm in a good mood so I will give hint for (b): Hint for (b): how many numbers end with 2,4,6, or 8? Figure out how many there are, then the probability is that number divided by the answer from (a). If you look at the last digit of the 3 digit number, there are C(4,1) ways to select an even digit. Once the ending digiit is known to be even, the other two digits can be even or odd. So C(4,1) must be multiplied by P(6,2) which is the number of ways of arranging the other 2 digits out of the 6 remaining digits. ==== For part (b), the value 1/6 is incorrect. The correct answer is 4/7 (=120/210). Also notice that there are 4 even and 3 odd digits, so 4/7 are even. Say you pick the ending digit first, that has a 4/7 chance of being even. So simple! Answer by Edwin McCravy(19211)     (Show Source): You can put this solution on YOUR website! a) How many three digit numbers can be formed from the digits 2,3,4,5,6,7 and 8? No digit repeats in the same number. There are 7 digits. Choose the 1st digit any of the 7 digits. That's 7 ways. Having chosen 1 digit, and since no digits can be repeated, there remain only 6 digits to choose from. Choose the 2nd digit any of the remaining 6 digits. That's 6 ways. Having chosen 2 digits, and since no digits can be repeated, there remain only 5 digits to choose from. Choose the 3rd digit any of the remaining 5 digits. That's 5 ways. That's 7×6×5 = 7P3 = 210 b) What is the probability of an even number? We first answer the question: "How many even three-digit numbers can we have. We choose the most restrictive thing first, which is the 3rd digit, for it must be even. Choose the 3rd digit any of the 4 even digits 2,4,6,8. That's 4 ways. Having chosen 1 digit, and since no digits can be repeated, there remain only 6 digits to choose from. Choose the 1st digit any of the remaining 6 digits. That's 6 ways. Having chosen 2 digits, and since no digits can be repeated, there remain only 5 digits to choose from. Choose the 2nd digit any of the remaining 5 digits. That's 5 ways. So that's 4×6×5 = 120 ways So the probability of an even number is 120 out of 720 or 120/720 which reduces to 1/6. c) What is the probability of numbers greater than 550 but less than 770? The first digit is either 5, 6 or 7, but we must take each of those as a separate case because with each, the number of choices for the 2nd digit changes. Case 1. the first digit is 5. Choose the first digit as 5. That's 1 way. Choose the 2nd digit any of the 3 digits 6,7,8. That's 3 ways. Having chosen 2 digits, and since no digits can be repeated, there remain only 5 digits to choose from for the 3rd digit. That's 1×3×5 = 15 ways for case 1. Case 2. the first digit is 6. Choose the first digit as 6. That's 1 way. Choose the 2nd digit as any of the 6 digits 2,3,4,5,7,8. That's 6 ways. Having chosen 2 digits, and since no digits can be repeated, there remain only 5 digits to choose from for the 3rd digit. That's 1×6×5 = 30 ways for case 2. Case 3. the first digit is 7. Choose the first digit as 7. That's 1 way Choose the 2nd digit any of the 5 digits 2,3,4,5,6. That's 5 ways. Having chosen 2 digits, and since no digits can be repeated, there remain only 5 digits to choose from. That's 5 ways. That's 1×5×5 = 25 ways for case 3. So for all three cases, that's a grand total of 15+30+25 = 70 ways So the probability of a number bettween 550 and 770 is 70 ways out of 210 or 70/210 which reduces to 1/3. Edwin

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