Question

The number of different seven-digit numbers that can be written using only three digits 1, 2 & 3 under the condition that the digit two occurs exactly twice in each number is

• A. $672$
• B. $640$
• C. $512$
• D. None of these

Answer 1

The correct option is A $672$

These 5 digits has to be a mixture of 1 and 2.
No of ways in which this can be possible :
When we calculate permutations remember these 2 cases
a) When no letter/number repeats and word has a length n : Total permutations = n!
b) When repetitions are there (example : one number repeats two times, second number repeats 9 times, total word length is n (n>10), in this case permutations will be = n!/[2! x 9!])
1) Using 1 five times
No of such cases (Two 3s, Five 1s, Zero 2s) : 7! / (2! x 5! x 0!)
2) 1 four times , 2 1 time
No of such cases (Two 3s, Four 1s, One 2s) : 7! / (2! x 4! x 1!)
3) 1 three times, 2 two times
No of such cases (Two 3s, Three 1s, Two 2s) : 7! / (2! x 3! x 2!)
4) 1 two times , 2 three times
No of such cases (Two 3s, Two 1s, Three 2s) : 7! / (2! x 2! x 3!)
5) 1 one time, 2 four times
No of such cases (Two 3s, One 1s, Four 2s) : 7! / (2! x 1! x 4!)
6) Using 2 five times
No of such cases (Two 3s, Zero 1s, Five 2s) : 7! / (2! x 0! x 5!)
No of Total cases :
(Add 1 to 6 case totals , Note * : Cases 1-3 and 4-6 are computationally same, so just multiply case 1-3 by 2)
7! x [ (1/120) + (1/24) + (1/12) ] = 7! x [ (1+5+10)/120] = 7! x [16/120]
= 7! x 16/5! = 16 x 7 x 6 = 672
Another (& better) approach to solve the given problem:
Assume 7 blanks are to be filled with 1,2,3 and 3 can be used only twice
so for each of the remaining 5 spaces we have 2 choices 1 and 2:
No of such cases : 7C2 x 2^5 = (7 x 6 x 32)/2 = 7 x 6 x16 = 672