how many three digit numbers can be made with the digits 1,2,3,4,5,6 and 7 if a) repetitions are allowed b) no digits is to be repeated in a number?
More
1 Expert Answer
Hi Thea
With 7 numbers, we can create three digit numbers like this.
a) With repetitions would suggest with replacement.
We'd have 7 options for the first digit, 7 options for the 2nd digit and 7 choices for the 3rd digit
73 = 343 combinations
b) Without repeated numbers, would suggest without replacement
We'd have 7 options for the first digit, 6 options for the 2nd digit and 5 choices for the 3rd
digit
7•6•5 = 210 combinations
Still looking for help? Get the right answer, fast.
OR
Find an Online Tutor Now
Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.
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:
Nội dung chính Show
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?
(ii) repetition of the digits is not allowed?
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?
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?
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?
Problem 5: A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?
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?
How many two digit numbers can be formed from 3/5 and 7 if no digit is repeated in any number?
How many 2 digit numbers can be formed with the digits 3 5 6 7 8 if none of the digits is repeated in any of the numbers formed?
How many 2 digit and 3 digit numbers can be formed using the digits 3 5 6 without repeating any digit?
How many 3 digits number can be formed using the digits 1 3 5 7 9 where we are allowed to repeat the digits?
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.
Answer : 64
Solution : One-digit numbers: <br> Clearly, there are four 1 -digit numbers. <br> Two-digit numbers: <br> We may fill the unit's place by any of the four given digits. <br> Thus, there are 4 ways to fill the unit's place. <br> The ten's place may now be filled by any of the remaining three digits. So, there are 3 ways to fill the ten's
place. <br> Number of 2-digit numbers `=(4xx3)=12.` <br> Three-digit numbers: <br> Number of ways to fill the unit's, ten's and hundred's places are 4, 3 and 2 respectively. <br> Number of 3-digit numbers `=(4xx3xx2)=24.` <br> Four-digit numbers: <br> Number of ways to fill the unit's ten's, hundred's and thousand's places are 4, 3, 2 and 1 respectively. <br> Number of 4-digit numbers `=(4xx3xx2xx1)=24.` <br> Hence, the number of required numbers
`=(4+12+24+24)=64.`
How many two digit numbers can be formed from 3/5 and 7 if no digit is repeated in any number?
Detailed Solution ⇒ Number of possible two-digit numbers which can be formed by using the digits 3, 5 and 7 = 3 × 3. ∴ 9 possible two-digit numbers can be formed.
How many 2 digit numbers can be formed with the digits 3 5 6 7 8 if none of the digits is repeated in any of the numbers formed?
∴ Required number of numbers = (1 x 5 x 4) = 20. Was this answer helpful?
How many 2 digit and 3 digit
numbers can be formed using the digits 3 5 6 without repeating any digit?
Solution : 357, 375, 537, 573, 735, 753. Therefore, '6' three-digit numbers can be formed.
How many 3 digits number can be formed using the digits 1 3 5 7 9 where we are allowed to repeat the digits?
Therefore, a total of 100 3 digit numbers can be formed using the
digits 0, 1, 3, 5, 7 when repetition is allowed.
How many 3
There is 4 possible ways to fill hundredth place as digits cannot be repeated. ∴ 3 - digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9 is 20.
How many 3
∴ Total number of 3-digit numbers = 3×4×5=60.
How many 3
Note: The number of 3-digit numbers formed using the digits 1, 2, 3, 4 and 5 when repetition is not allowed is equivalent to the number of 3 Letter permutations of 5 distinct letter = 5P3=5! (5−3)! =5!2!
How many 3
Answer: 60
As repetition is not allowed, So the number of digits available for B = 4 (As one digit has already been chosen at A), Similarly, the number of digits available for C = 3. Thus, The total number of 3-digit numbers that can be formed = 5 × 4 × 3 = 60.
