# Permutation question?

Find the number of 4 digit figures that can be formed using 4 digits which lie between 3500 and 6000 from the numbers 2; 3; 4; 5; 6; 7; 8; 9 without repetition.

Relevance

2*7*6*5 + 5*6*5 = 570

• Anonymous
9 years ago

From 2 to 9 there are total "8" numbers.

First find all numbers between 3500 and 4000

Make four boxes to allot numbers,

In first box you can put 3 in "1" way

In second box you can put 5,6,7,8,9 in "5" ways..................6 no. remains

In 3rd and 4th box you can put remaining 6 numbers in 6P2 ways

Total no. of ways = 1 x 5 x 6P2 = 150

Now, find all numbers between 4000 and 6000

In first box you can put 4 and 5 in "2" way...................7 no. remains

In 3 boxes you can put remaining 7 numbers in 7P3 ways

Total no. of ways = 2 x 7P3 = 420

Total numbers b/w 3500 and 6000

= 150 + 420

= 570

• Anonymous
9 years ago

1*5*6P2 + 1*7P3 + 1*7P3

@ kashish : first digit can be 3 or 4 or 5.

if the first digit is 3 then the second digit has to be greater than or equal to 5 cuz minimum nuber is 3500 so second digit can be 5,6,7,8,9

so, 1*5*6P2 ways

if first digit is 4 or 5 then other digits can be arranged is 7P3 ways.

so 1*7P3 + 1*7P3.

so total 1*5*6P2 + 1*7P3 + 1*7P3 = 570 ... answer