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Find 8P6
Find 8C6
Find the number of ways in which the letters of the word BIHAR can be arranged.
Since there are 5 letters in the word so the total number of arrangement is 5P5 = 5! that is 120.
Find the number of ways in which the letters of the word AMERICA can be rearranged.
In the word AMERICA there are 7 letters and the letter ‘A’ is coming twice.
So the total number of rearrangement
Note: No. of Rearrangements = Total no. of arrangements – 1(initial arrangement)
Find the number of ways in which the letters of the word CALCUTTA can be rearranged.
In the word CALCUTTA, there are 8 letters
Letter C comes twice
Letter A comes twice
Letter T comes twice
So the total number of rearrangements
Note: No. of Rearrangements = Total no. of arrangements – 1 (initial arrangement)
How many 4-letter different words can be formed by using the letters a, b, c and d?
Letter a can be placed in all 4 positions.
Similarly, b can be placed in all 4 positions.
Similarly, c can be placed in all 4 positions.
Similarly, d can be placed in all 4 positions.
So the total number of arrangement is 4 × 4 × 4 × 4 = 256
How many numbers can be formed by using the digits 2, 3, 4 and 5?(repetitions of digits is not allowed).
The number formed can be one-digit, two-digit, three-digit or four-digit number
Number of one-digit numbers formed = 4
Hence, total numbers formed = 4 + 12 + 24 + 24 = 64
How many numbers greater than 4000 can be formed by using the digits 2, 3, 4 and 5? (repetitions of digits is not allowed).
Thousand’s place can be filled by using two digits i.e. 4 or 5
Hundred’s place can be filled in 3 ways.
Ten’s place can’t be filled in 2 ways.
Unit’s place can be filled in 1 way.
So total number of possible numbers = 2 × 3 × 2 × 1 = 12.
How many numbers greater than 4000 can be formed by using the digits 2, 3, 4 and 5?
Thousands place can be filled in two ways.
Hundred’s place can be filled with any of the 4 digits.
Ten’s place can be filled with any of the 4 digits.
Unit’s place can be filled with any of the 4 digits.
So the total numbers formed = 2 × 4 × 4 × 4 = 128.
There are 3 roads from A to B, 4 roads from B to C and only 1 road from C to D. How many combinations of roads are there from A to D?
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