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: Show
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(ii) repetition of the digits is not allowed?Solution:
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:
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:
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:
Problem 5: A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?Solution:
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 : 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.` 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. ∴ Required number of numbers = (1 x 5 x 4) = 20. Was this answer helpful? Solution : 357, 375, 537, 573, 735, 753. Therefore, '6' three-digit numbers can be formed. Therefore, a total of 100 3 digit numbers can be formed using the digits 0, 1, 3, 5, 7 when repetition is allowed.
Solution: A two-digit number has 2 digits, one at the units place and one at the tens place. Let's form the two-digit numbers using the digits 3, 7, and 9 if the repetition of the digit is allowed. Below are all the numbers formed using the digits 3, 7, 9 with 3 at tens place: 33, 37, 39 Below are all the numbers formed using the digits 3, 7, 9 with 7 at tens place: 73, 77, 79 Below are all the numbers formed using the digits 3, 7, 9 with 9 at tens place: 93, 97, 99 Therefore, the two-digit numbers formed by using the digits 3, 7, and 9 are 33, 37, 39, 73, 77, 79, 93, 97, and 99. Summary: All possible two-digit numbers formed by using the digits 3, 7, and 9 if repetition of the digit is allowed are 33, 37, 39, 73, 77, 79, 93, 97, and 99.
Answer: 12 ways Step-by-step explanation: Using the permutation formula P(n,r) = n! / (n-r)! P(4,2) = 4! / (4-2)! P(4,2) = 4! / 2! P(4,2) = 4×3×2×1 / 2×1 P(4,2) = 4×3 P(4,2) = 12 ways |