In how many ways one can arrange 10 letters taken 7 at a time. In how many of these 3 particular letters (i) always occur, and (ii) never occurs?
Answers
One can arrange 10 letters taken 7 at a time in 6,04,800 ways.
(i) Number of arrangements in which 3 particular letters always occur is 5880
(ii) Number of arrangements in which 3 particular letters never occur is 5040
Given:
Number of letters = 10
Number of letters taken at a time = 7
To find:
(a) Number of ways one can arrange 10 letters taking 7 at a time
(b) The number of arrangements in which 3 particular letters (i) always occur (ii) never occur
Solution:
(a) The formula for permutation of n object taken r at a time is given by
P(n,r) =
According to the given data,
n = 10
r = 7
The number of ways to arrange 10 letters taking 7 at a time
= P(10,7)
= = 6,04,800
(b) (i) Here, 3 particular letters always occur. So we fix 3 places in the arrangement for those letters.
Number of letters remaining = 10 - 3 = 7
Number of letters that can be taken at a time = 7 - 3 = 4
Using the formula of conditional permutations, the number of arrangements possible
= 7 x P(7,4)
= 7 x (7! / (7 - 4)!)
= 7 x (7! / 3!)
= 7 x 840
= 5880
(ii) Here, 3 particular letters never occur. So, we remove those letters from the total number of letters.
Number of letters remaining = 10 - 3 = 7
Number of letters to be taken at a time = 7
The number of arrangements possible
= P(7,7)
= 7! / (7-7)!
= 5040
Hence, one can arrange 10 letters taken 7 at a time in 6,04,800 ways.
(i) Number of arrangements in which 3 particular letters always occur is 5880
(ii) Number of arrangements in which 3 particular letters never occur is 5040
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