Total Questions: 51
When A or E are first or fourth letter then arrangement = 2! × 3! = 12When A or E are second or fifth letter then aggangement = 2! × 3! = 12Hence, total arrangement = 12 + 12 = 24
Number of different way from A to C =3 + 3 × 4 = 15
Hence, maximum number of attempts required = 6 + 3 + 1 = 10
Hence, total different password = 5 × 9 × 21 = 945
∴ Number of arrangement = 3 × 3 = 9Again cups in 3 row arrange in 3! = 6 way∴ Number of ways = 9 × 6 = 54And cups in 3 row are arrange in 3 × 3 × 3 = 27 waysHence, total number of ways = 54 + 27 = 81
1. It is possible that exactly one litter goes into an incorrect envelope.2. There are only six ways in which only two letters can go into the correct envelopes.
Which of the statements given above is/are correct?
Total number of ways of putting all four letters into four envelope = 4! = 24Suppose A, B, C and D are 4 envelope and L1, L2, L3 and L4 are four letters with correct address respectively.S1: When L1 is placed in either of envelope B, C or D into an incorrect address, then placed into envelope A into an incorrect address.Hence, it is not possible that exactly one letter goes into an incorrect envelope.Hence, Statement 1 is not correct.S2: Possible ways in which only two letters can go into correct address.
Only 6 possible ways in which only two letters can go into correct envelopes.Hence, Statement 2 is correct.
Number of ways of arrangement of odd digits
at odd place = 4! / (2! × 2!) = 3 × 2 = 6
Number of ways of arrangement of even digits atEven place = 4! / (2! × 2!) = 3 × 2 = 6.
Total Arrangement = 6 × 6 = 36