+0  
 
0
83
1
avatar

How many distinct odd-digit numbers can be written with the digits 1,2,3, 4 and 5 if no digit may be used more than once?

 Jun 14, 2021
 #1
avatar+524 
+2

Digits are 1, 2, 3, 4 and 5. 

⇒no. of 1-digit odd no.'s  = 3

           [Repetition of digits not allowed] 

 

No. of 2-digit odd numbers

          one's place has = 3 possibilities 

          Then, 10's place has = 4 possibilities

⇒ no. of 2-digit odd numbers = 4 × 3  = 12 

 

No. of 3-digit odd numbers

           one's place has = 3 possibilities 

           10's place has = 4 possibilities 

           then, 100's place has = 5 - 2 = 3 possibilities 

 ⇒ no. of 3-digit odd numbers = 3 × 4 × 3 = 36 

 

No. of 4-digit odd numbers

           one's place has = 3 possibilities

           10's place has = 4 possibilities

           100's place has = 3 possibilities

           then, 1000's place has = 2 possibilities

⇒ no. of 4-digit odd numbers = 3 × 4 × 3 × 2 = 72

 

No. of 5-digit odd numbers

           one's place has = 3 possibilities

           10's place has = 4 possibilities

           100's place has = 3 possibilities

           1000's place has = 2 possibilities

           then, 10000's place has = 1 possibility

⇒ no. of 5-digit odd numbers = 3 × 4 × 3 × 2 × 1 = 72

 

∴ No. of odd numbers = 72 + 72 + 36 + 12 + 3 = 195

 Jun 14, 2021

13 Online Users