The Railway Recruitment Board has released the RRB Group D Answer Key on 14th October 2022. The candidates will be able to raise objections from 15th to 19th October 2022. The exam was conducted from 17th August to 11th October 2022. The RRB (Railway Recruitment Board) is conducting the RRB Group D exam to recruit various posts of Track Maintainer, Helper/Assistant in various technical departments like Electrical, Mechanical, S&T, etc. The selection process for these posts includes 4 phases- Computer Based Test Physical Efficiency Test, Document Verification, and Medical Test. 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?Solution: Índice
(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.
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 |