if the 6 digit numbers formed from the digits 1,2,3,4,5,6 are arranged in the increasing order(repetition not allowed),then 291th term will be?

Question

if the 6 digit numbers formed from the digits 1,2,3,4,5,6 are arranged in the increasing order(repetition not allowed),then 291th term will be?

Answer 1

Arthur D. gave an excellent, complete math answer.  This is some more detail to help you understand the process (PLZ don't let it confuse you):

 

The 291st term is 341526

This is because both the decimal (that is, base 10) and the quinary (that is, base 5 – even though these digits are 1-6 instead of 0-5) are positional number systems.

The digits in the ones place go from 1 to 6. These are always less than the digits in the fives (not tens) place that go from 1 to 6, which are always less than the digits in the twenty-fives place, etc.

Since repetition is not allowed, you count by putting each of the remaining digits in the ones place (1 choice), then put each of the remaining digits in the fives place (2 choices), then put each of the remaining digits in the twenty-fives place (1*2*3=6 choices) … until you reach or exceed 291.

It is extremely helpful to know how many digits are needed before 291 is exceeded. The formula for permutations without replacement, P(n,r) = n!/(n-r)!, is essential. When n=6, how large does r have to be to get a value for P that is roughly 291?
             Important: for n=6, r=4, P(6,4) = 6*5*4*3*2*1 / 2*1 = 6*5*4*3 = 360

                              for n=6, r=3, P(6,3) = 6*5*4*3*2*1 / 3*2*1 = 6*5*4 = 120

If we start with 1xxxxx, we can count 120 terms from 123456 to 165432.
There are 120 more terms from 213456 to 265431.        240 so far, so we continue with 312xxx.
Now, groups of P(6,2)= 30 [1 group gets us to 270].      We continue with 324xxx.

                           [no repetition, so 321xxx is followed by 324xxx]

Then, groups of P(6,1) = 6 [this will take (291-270)/6 = 3 remainder 3]. Three groups are:
       324156 to 324651       [271-276]

       325146 to 325641       [277-282]

       326145 to 326541      [283-288]

       341256               [289]              [3 more for remainder 3 now]

       341265               [290]

       341526               [291]

So we have 341526.

We can count:
  1   123456 123345x has 1 choice
  2   123465 and 12346 has 1 choice, but 1234xx has a total of 2 choices
  3   123546
  4   123564 and 1235xx also has 2 choices
  5   123645
  6   123654 but 123xxx has a total of 6 choices
  7   124356
  8   124365
  9   124536
 10   124563
 11   124635
 12   124653 but 124xxx has 6 more choices
 13   125364
. . .

 24   126543 the last of the 126xxx terms

 25   132456 the last of the 13xxxx terms (because reuse is not allowed)
. . .

 49   142356 the last of the 14xxxx terms
. . .

 96   156432 the last of the 15xxxx terms
 97   162345
. . .
288    326541 the last of the 32xxxx terms (no 33xxxx terms because no repetition)
289    341256
290    341265
291    341526     *** This is term 291.
. . .

 

{note: I verified this using a very short, easy computer program.]

Answer 2

there are 6!=6*5*4*3*2*1=720 numbers altogether without repetitions

there are 5!=5*4*3*2*1=120 numbers that start with the number 1

there are 5!=5*4*3*2*1=120 numbers that start with the number 2

that is a total of 240 numbers so far

the next 120 start with the number 3

these are the 241st through the 360th numbers

these numbers can begin with 31, 32, 34, 35, and 36

there are 120/5=24 of each of these sets of numbers

31XXXX  (24 numbers)

32XXXX  (24 numbers)

34XXXX  (24 numbers

35XXXX  (24 numbers)

36XXXX  (24 numbers)

we have 240 numbers so far

use up all the 31XXXX numbers and all the 32XXXX numbers

that gives us 240+24+24=288 numbers

34XXXX gives the 289th through the 312th numbers

you want the 291st number

341256 is the 289th number

341265 is the 290th number

341,526 is the 291st number

Answer 3

291 divided by 6 has a remainder of 3, which means the 3rd term, which is coincidentally a 3, will be in spot 291.