General Questions
Practice and master this topic with our carefully crafted questions.
The word 'OPTICAL' contains 7 different letters.
When the vowels OIA are always together, they can be supposed to form one letter.
Then, we have to arrange the letters PTCL (OIA).
Now, 5 letters can be arranged in 5! = 120 ways.
The vowels (OIA) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6) = 720.
We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).
Required number of ways
| = (3C1 x 6C2) + (3C2 x 6C1) + (3C3) | ||||||||||||||
|
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| = (45 + 18 + 1) | ||||||||||||||
| = 64. |
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.
'LOGARITHMS' contains 10 different letters.
Required number of words = Number of arrangements of 10 letters, taking 4 at a time.
| = 10P4 | |
| = (10 x 9 x 8 x 7) | |
| = 5040. |
In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.
Thus, we have MTHMTCS (AEAI).
Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.
Number of ways of arranging these letters :
| 8! | = 10080. |
| (2!)(2!) |
Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.
Number of ways of arranging these letters :
| 4! | = 12. |
| 2! |
Required number of words = (10080 x 12) = 120960.
The word ‘BULLET’ contains 6 letters of which 1 letter occurs twice = 6! / 2! = 360
No. of permutations possible with vowels always together = 5! * 2! / 2! = 120
No. of permutations possible with vowels never together = 360-120 = 240.
In order that two books on Hindi are never together, we must place all these books as under:
H E H E H E H….. H E H
Where H denotes the position of Hindi book and E that of English book.
Since there are 22 books on English, the number of places marked E are 23.
Now, 20 places out of 23 can be chosen in 23C20
| = (23C3 | ||||||
|
||||||
| = 1771. |
Hence the number of ways = 1771 ways
As A and T should occupy the first and last position, the first and last position can be filled in only one way. The remaining 4 positions can be filled in 4! Ways by the remaining words (S,C,E,N,T).
Hence by rearranging the letters of the word ASCENT we can form 1x4! = 4! words.
Circular Permutation
'n' objects can be arranged around a circle in (n1)!
If arranging these 'n' objects clockwise or counter clockwise means one and the same, then the number arrangements will be half that number.
| i.e., number of arrangements = | (n - 1)! |
| 2 |
Let there be exactly one person between the two brothers as stated in the question.
If we consider the two brothers and the person in between the brothers as a block, then there will 17 others and this block of three people to be arranged around a circle.
The number of ways of arranging 18 objects around a circle is in 17! ways.
Now the brothers can be arranged on either side of the person who is in between the brothers in 2! ways.
The person who sits in between the two brothers could be any of the 18 in the group and can be selected in 18 ways.
Therefore, the total number of ways 18 × 17! × 2 = 2 × 18!
There are 8 students and the maximum capacity of the cars together is 9.
We may divide the 8 students as follows:
Case I: 5 students in the first car and 3 in the second
Case II: 4 students in the first car and 4 in the second
Hence, in Case I: 8 students are divided into groups of 5 and 3 in 8C3 ways.
Similarly, in Case II: 8 students are divided into two groups of 4 and 4 in 8C4 ways.
Therefore, the total number of ways in which 8 students can travel is:
(8C3 + 8C4) = 70 + 56 = 126