The number of arrangements of 4 different digits taken 4 at a time is given by = 4! = 24.All the four digits will occur equal number of times at each of the position,namely ones,tens,hundreds,thousands.
Thus,each digit will occur 24/4 = 6 times in each of the position.The sum of digits in one's position will be 6 x (1+3+5+7) = 96.Similar is the case in ten's,hundred's and thousand's places.
Therefore,the sum will be 96 + 96 x 10 + 96 x 100 + 96 x 100 = 106656
We have to arrange 6 books.
The number of permutations of n objects is n! = n. (n ? 1) . (n ? 2) ... 2.1
Here n = 6 and therefore, number of permutations is 6.5.4.3.2.1 = 720
Here in 100P2, P says that permutations and is defined as in how many ways 2 objects can be selected from 100 and can be arranged.
That can be done as,
= 100!/(100 - 2)!
= 100 x 99 x 98!/98!
= 100 x 99
= 9900.
Given word is TRANSFORMER.
Number of letters in the given word = 11 (3 R's)
Required, number of ways the letters of the word 'TRANSFORMER' can be arranged such that 'N' and 'S' always come together is
10! x 2!/3!
= 3628800 x 2/6
= 1209600
In a 3 digit number one?s place can be filled in 5 different ways with (0,2,4,6,8)
10?s place can be filled in 10 different ways
100?s place can be filled in 9 different ways
There fore total number of ways = 5X10X9 = 450
Given word is THERAPY.
Number of letters in the given word = 7
These 7 letters can be arranged in 7! ways.
Number of vowels in the given word = 2 (E, A)
The number of ways of arrangement in which vowels come together is 6! x 2! ways
Hence, the required number of ways can the letters of the word 'THERAPY' be arranged so that the vowels never come together = 7! - (6! x 2!) ways = 5040 - 1440 = 3600 ways.
NUMERICAL has 9 positions in which 2, 4, 6, 8 are even positions.
And it contains 5 consonents i.e, N, M, R, C & L. Hence this cannot be done as 5 letters cannot be placed in 4 positions.
Therefore, Can't be determined.
There are total 9 places out of which 4 are even and rest 5 places are odd.
4 women can be arranged at 4 even places in 4! ways.
and 5 men can be placed in remaining 5 places in 5! ways.
Hence, the required number of permutations = 4! x 5! = 24 x 120 = 2880
First letter can be posted in 4 letter boxes in 4 ways. Similarly second letter can be posted in 4 letter boxes in 4 ways and so on.
Hence all the 5 letters can be posted in = 4 x 4 x 4 x 4 x 4 = 1024
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