Permutation & combination

Question

Q: how many no can be formed using 1,2,3,4,5,6,7,8,9 digits and without repeating the digits?

Conditions are:1) it should be 4 digit no.

                     2) no should be odd

                    3) 9 should be used exactly once

Please expain

Comments

from condition 3 we can conclude that digits in 4 digit number can be repeated except 9 right ??

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 deep_downYour answer for without repetition is right.

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i have updated the q that 9 should be used exactly once and no repetition

Answer 1

Numbers ={1,2,3,4,5,6,7,8,9}

4 digits _ _ _ _

(A)With Repetition

 

Case(1)unit place digit is 9.

 

According to 3rd condition 9 should be used exactly once.If we choose 9 for unit place then Number of choices for remaining places=8 each  

Therefore no.of 4 digit numbers that can be formed= 1*8*8*8=512

 

Case(2) Unit place filled with 1,3,5,7 

 

According to second condition number Should be odd.So,there are 5(1,3,5,7) choices for unit place.

 

Now for digit 9 there are 3 choices (2nd place or 3rd place or 4th place).and for remaining two Number of choice=8 

 

Therefore no.of 4 digit numbers that can be formed

= 4*1*8*8 + 4*8*1*8 +4*8*8*1=768

 

Total 4 digit  number that can be formed with repetition= 512 +768 =1280

(B)Without Repetition

Case(1)when unit place digit is 9

Number of Choice for unit Place =1

Number of Choice for 2nd Place =8

Number of Choice for 3rd Place =7

Number of Choice for 4th Place =6

Therefore no.of 4 digit numbers that can be formed= 1*8*7*6=336

Case(2)When Unit Place filled with 1,3,5,7.

Number of Choices for unit Place =4

Here 9 can come at 2nd or 3rd or 4th place.

 

Therefore no.of 4 digit numbers that can be formed

= 4*8*7*6=1344

 

Total 4 digit number that can be formed without repetition= 336 +1344 =1680

 

Answer 2

From condition 3, it is clear that digits in the 4 digit number can be repeated except digit 9.

Now, 4 digit no should be odd, so unit digit can be among 1, 3, 5, 7, 9.

case 1- If 9 is at unit place.

     4 digit no = 8 * 8 * 8 * 1 = 512.

case 2 - If 9 is not at unit place, then unit place digit should be among 1, 3, 5, 7. And 9 must be there at tens place or at hundreds place or at thousands place.  

     4 digit no = 3 * 8 * 8 * 4 = 768

So Total 4 digit no. = 512 + 768 

                             = 1280 .(Ans)

Answer 3

Number should be in between 1,3,5,7,9

where 9 is possible exactly once.

Now for 4 digit number can choose 9   ⁸C₁  ⁸C₁  ⁴C₁ =8⨉8⨉4 ways =256 ways

9 in 2nd position "       "         "       ⁸C₁  9 ⁸C₁  ⁴C₁   = 256 ways

9 in 3rd position  "       "        "       ⁸C₁  ⁸C₁ 9   ⁴C₁ =256 ways

9 in 4th position "        "         "      ⁸C₁   ⁸C₁  ⁸C₁  9  =512 ways

So, total numbers 256+256+256+512=1280

  

Comments

i think ans for without repetion & 9 is used exactly once :

9 can be placed at unit, hundred,thousand & ten thousand place so there are 4 possibilty

1. when 9 is at unit place: 8*7*6*1=336

2.when 9 is at tens place: 7*6*1*4-=168

3.when 9 is at hundred place: 7*1*6*4=168

4.when 9 at thousand place: 1*7*6*4= 168

so ans will be 336+168+168+168=840

above answers are right if there is no restrictions over 9 but 9 should be used exactly once it means 9 should be there is in the no .