Permutations

1)In how many ways can the letter of the word 'ARRANGE' be arranged so that

a)the 2 A's are together but not the 2 R's ?

b)neither the two A's nor the two R's are together?

2)How many numbers greater than 10000 can be formed from the digits of the numbers 112340?he numbers may be either of 5 digits or 6 digits.

Am getting the answer as 456, which is incorrect [2]

9 Answers

62
Lokesh Verma ·

no of words in which A's are together

minus

no of words in which A's and R's are together..

= 6!/2 - 5! = 2. 5!

62
Lokesh Verma ·

2)How many numbers greater than 10000 can be formed from the digits of the numbers 112340?he numbers may be either of 5 digits or 6 digits.

all that you should do is make a 6 digit number and allow 0 to be the first digit [1]

so the answer will be 6!/2! = 360

11
Khyati ·

Sir, 2nd question ka answer is given as 600. These questions are from KC Sinha.

39
Dr.House ·

2)

2 cases :

cas1: no :of 5 digit numbers we can make

sub case1:

those with 1 as the left most digit

then remaining 4 places can be filled in 5P4 ways =120

sub case 2: those with 2,3,4 as left most digit

so left most can be filled in 3 ways and remaining 4 places can be filled in 5P4/2! ways

so for sub case 2 , totla numbers =3x5P4/2!

so total no:of 5 digit numbers we can make = 120 +3x5P4/2! =300

case2: no of 6 digit numbers we can make

sub case 1: with 1 as left most digit

then remaining 5 digits can be filled in 5! ways = 120

sub case 2: with 2,3,4 as left most digit

then remaining 5 digits can be filled in 5!/2! ways

so total numbers =3x5!/2!=180

so total 6 digit numbers =300

so total numbers we can make>10,000 =300+300=600

39
Dr.House ·

and nishant sir

with reference to your comment

all that you should do is make a 6 digit number and allow 0 to be the first digit

then in that case u only can calculate all 5 digit numbers >10000

u will be missing the 6 digit numbers

and coming to your calculation 6!/2! , that is permuattion of 6 things in 6 places when 2 are alike

that doesnt complement your above comment in anyway :P

62
Lokesh Verma ·

I did not say that you have to force 0 as the first digit..

what i tried to imply was that if 0 is teh first digit... nothting to worry.. it will give us a 5 digit number!

figured out my mistake..

i missed those 5 digit numbers where 0 is a part of the number..

39
Dr.House ·

exactly that wat i was going to say

while doing 6!/2! u missed those 5 digit numbers of which 0 is a part

11
Khyati ·

Thanks Nishant Sir and id no 711.(kitne naam change karte ho aap [3])

11
Khyati ·

1)In an election the number of candidates is one more than the number of members to be elected. If a voter can vote in 30 different ways, find the number of candidates.(A voter has to vote for at least one candidate)

2)In a plane there are n lines, no two of which are parallel and no three are concurrent. How many triangles can be formed with their points of intersection as vertices?(Please dont make use of binomial theorem while solving this one, because I wanna know the method by which this kinda sums can be solved using PnC concepts only)

3)There is a convex polygon of n sides (n>5). Triangles are formed by joining the vertices of the polygon,how many triangles are there and how many of them have no side side common with any of the sides of the polygon.

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