Ex 6.3, 10
In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?
Total number of permutation of 4I not coming together
= Total permutation
– Total permutation of I coming together
Total Permutations
In MISSISSIPPI
there are 4I, 4S, 2P and 1M
Since letters are repeating,
we will use the formula = 𝑛!/𝑝1!𝑝2!𝑝3!
Total number of alphabet = 11
Hence n = 11,
Also, there are 4I, 4S, 2P
p1 = 4, p2 = 4, p3 = 2
Hence,
Total number of permutations = 𝑛!/𝑝1!𝑝2!𝑝3!
= 11!/(4! 4! 2!)
= (11 × 10 × 9 × 8 × 7 × 6 × 5 × 4!)/((4 × 3 × 2 × 1) (4!)×(2 × 1))
= 34650
Total permutations of I coming together
Now taking 4Is as one,
MISSISSIPPI
Here, there are repeating letters
So, we use the formula , Number of permutation = 𝑛!/𝑝1!𝑝2!
Number of letters = 8
∴ n = 8
Since there are 4 S & 2 P
p1 = 4, p2 = 2,
Number of permutation with 4I together = 𝑛!/𝑝1!𝑝2!
= 8!/(4! 2!)
= 840
Now,
Total number of permutation of 4I not coming together
= Total permutation – total permutation of I coming together
= 34650 – 840
= 33810
Made by
Davneet Singh
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo
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