The number of boxes containing same number of oranges will be least when there are maximum boxes containing different number of oranges.
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The number can end in multiples of 4, that is 12, 16, 24, 36, 32, 52, 56, 64 = 8 cases. The first three positions can be filled by 4 × 3 × 2 = 24 ways. Hence total number of ways = 24 × 8 = 192 ways. |
Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
Required number of numbers = (1 x 5 x 4) = 20.
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Sol. In the given word, we treat the vowels IEO as one letter. Thus, we have DRCTRY (IEO). This group has 7 letters of which R occurs 2 times and others are different. Number of ways of arranging these letters = 7! X 3! /2! = 15120 3 vowels can be arranged among themselves in 3! = 6 ways |
Suppose first toss is A, second is B. We know that p(A_heads) = 50% and that p(B_tails) = 50%. Also, A and B are independent. So, p(A_heads and B_tails) = p(A_heads) * p(B_tails) = 50% * 50% = 25% = 1/4. Answer is C.