Question

The number of different seven-digit numbers that can be written using only three digits 1,2 and 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

Hint:
Here, we will find the number of ways of the arrangement of the digit if two occur exactly twice in each seven-digit number by using the combination formula. Then by using the powers, we will find the number of ways of arranging the remaining five-digit number. We will then multiply the number of the arrangement of digits in both the cases to get the number of different seven-digit numbers.

Formula Used:
We will use the following formula:
1) Combination is given by the formula $${}^n{C}_r={\displaystyle \frac{n!}{(n-r)!r!}}$$
2) Factorial is given by the formula $$n!=n\times \left(n-1\right)!$$

Complete step by step solution:
We are given the number of different seven-digit numbers that can be written using only three digits 1,2 and 3. Therefore,
Total number of Digits $$=7$$
We are given that the digit two occurs exactly twice in each number.
Thus, the digit two occurs twice in the seven digit number.
Now, we will find the number of ways of arrangement of the digit two in the seven digit number by using combination.
Total number of ways that the digit two occurs exactly twice in each number $$={}^7{C}_2$$
Now, the remaining five digits can be written using two digits 1 and 3 in $$2^5$$ ways.
We will now find the total number of seven digit number by multiplying the number of ways of arrangement in both the cases. Therefore
Total number of seven digit number $$={}^7{C}_2\times 2^5$$
Now by using the formula $${}^n{C}_r={\displaystyle \frac{n!}{(n-r)!r!}}$$, we get
$$\Rightarrow$$ Total number of seven digit number $$={\displaystyle \frac{7!}{\left(7-2\right)!2!}}\times 2^5$$

Subtracting the terms in the denominator, we get
$$\Rightarrow$$ Total number of seven digit number $$={\displaystyle \frac{7!}{5!2!}}\times 2^5$$
We know that the factorial can be written by the formula $$n!=n\times \left(n-1\right)!$$ , so we get
$$\Rightarrow$$ Total number of seven digit number $$={\displaystyle \frac{7\times 6\times 5!}{5!2!}}\times 2^5$$
$$\Rightarrow$$ Total number of seven digit number $$={\displaystyle \frac{7\times 6}{2!}}\times 2^5$$
Simplifying the expression, we get
$$\Rightarrow$$ Total number of seven digit number $$=7\times 6\times 2^4$$
Multiplying the terms, we get
$$\Rightarrow$$ Total number of seven digit number $$=672$$
Therefore, the number of different seven-digit numbers that can be written using only three digits 1,2 and 3 is 672.

Thus, option (A) is the correct answer.

Note:
We know that there is not much difference between permutation and combination. Permutation is the way or method of arranging numbers from a given set of numbers such that the order of arrangement matters. Whereas combination is the way of selecting items from a given set of items where order of selection doesn’t matter. Both the word combination and permutation is the way of arrangement. Here, we will not use permutation because the order of toys is not necessary.