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A permutation is often represented by the cycles it has, For example, if we permute the numbers in the natural order to 2 3 1 5 4, this is represented as (1 3 2) (5 4). In this the (132) says that the first number has gone to the position 3, the third number has gone to the position 2, and the second number has gone to the position 1, and (5 4) means that the fifth has gone to position 4 and fourth has gone to position 5. The numbers with brackets are to be read cyclically. If a number has not changed position, it is kept as a single cycle. This 5 2 1 3 4 is represented as (1345)(2). We may apply permutations on itself if we apply the permutation (132)(54) once, we get 2 3 1 5 4. If we consider the permutation of 7 numbers (1457)(263) what is its order ( how many times must it be applied before the numbers appear in the original order) ?
a. 12
b. 7
c. 7!
d, 14
Read Solution (Total 8)
-
- Ans. 12
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Explanation:
Since number of ways "n" can be arranged in a cyclic order= (n-1)!
Therefore,
(1457) can be arranged in = (4-1)! ways = 3! ways = 6 ways
&
(263) can be arranged in = (3-1)! ways = 2! ways =2
so, number of ways before the number appear in the original order= 6x2 = 12 ways. - 9 years agoHelpfull: Yes(21) No(0)
- 4*3=12 ans
- 10 years agoHelpfull: Yes(12) No(7)
- consider the cycle (1 4 5 7). If this is applied four times, the numbers 1, 4,
5 and 7 will be back in their original positions. The same will be true if you
apply it eight, twelve, sixteen or any multiple of four times.
For (2 6 3) this happens after three, six, nine, etc applications.
All seven numbers will be in their original positions if you apply the permutaion
(1 4 5 7) (2 6 3) a number of times that is a common multiple of 3 and 4. Find
the lowest common multiple.
Ans: 12
Source: Quora (https://www.quora.com/What-is-the-order-of-the-permutation-1-4-5-7-2-6-3) - 9 years agoHelpfull: Yes(5) No(0)
- factorial of 7 ans
- 9 years agoHelpfull: Yes(2) No(3)
- not yet solved
- 9 years agoHelpfull: Yes(1) No(5)
- can you xplain pulkit how??
- 9 years agoHelpfull: Yes(1) No(0)
- Since we have total no is 20
so remaining is 8
then
20-8=12 - 8 years agoHelpfull: Yes(1) No(3)
- for Numbers 1,2,3,4,5 if we permutate in (1 3 2) (5 4) order then it will take 3 cycles to get the original number.
1st time: 2 3 1 5 4
2nd time: 3 1 2 4 5
3rd time: 1 2 3 4 5.
if we permutate the 1 2 3 4 5 6 7 in (1 4 5 7) (2 6 3) order it will take 4 cycles to get the original position.
1st time: 7 3 6 1 4 2 5
2nd time: 5 6 2 7 1 3 4
3rd time: 4 2 3 5 7 6 1
4th time: 1 2 3 4 5 6 7.
So the answer should be 4times. - 7 years agoHelpfull: Yes(0) No(0)
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