Friday, 16 January 2015

Permutation

1. In how many ways can 6 persons be arranged at a random table so that 2 particular persons may sit together?

Ans:
 i)  Arrange with respect to table: Taking 2 particular persons as one person, we are to arrange 5 (=1+4) persons in 5!. Again 2 persons can be arranged among themselves in 2! ways.

So required no. of arrangement = 5! x 2! =120 x 2=240


ii) Arrangement with respect to each other : At first 2 particular persons can be arranged themselves in 2! ways. Keeping them fixed & taking   as one person, all the 5 persons can be arranged in (5-1)!=4! ways.

So required no. of ways=4! x 2!=24 x 2=48


2. The number of permutation of n different things taken r at a time in which p particular things never occur is
3. In how many of the permutations of 8 things taken 3 at a time, will two particular things never occur?

Ans: The no. of ways= (8-2)!/3!=120

4. Find the number of different numbers of 3 digits that can be formed with the digits 1,2,3,4,5 the digits in any number being all different and the unit place being always 5.

Ans: The required no. of ways= (5-1)!/(3-1)!=12

5. In how many ways can the letters of the word TABLE be arranged taken 4 at a time so that BL is always together and also in the order given.

Ans: The Required no. of ways =(4-2+1) x (5-2)!/(4-2)!=18

6. There are 10 stations in a railway line. How many different kinds of 2nd class tickets must be printed so as to enable a passenger to go from any station to another.

Ans: In the railway line there are 10 stations. So to travel from one particular station to any other station, 9 different types of tickets. Now for 10 stations , 10 x 9 =90 different types of tickets are necessary. 
So number of ways=90

7. In how many ways 5 (XI) students and 3 (XII) students be arranged in a row so that no (XII) students may be together?

Ans: 5 (XI) students may be arranged among themselves in 5! ways=120 ways.

In between 5(XI) students and at the extremities there are in all 6 places where 3 (XII) students may be placed in 6!/3!=120 ways.

So the total number of ways=120 x 120=14400   

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