How many 10 digit numbers can be formed using the digits 1 and 2?
A) $^{10}{P_2}$
B) $^{10}{C_2}$
C) ${2^{10}}$
D) $10!$
Answer
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Hint: We will first of all make 10 blank spaces which need to be filled using the digits 1 and 2. Then, we will write the possibility and ways to fill up each of the spaces individually and then multiply those ways to get the required answer.
Complete step-by-step answer:
We have to form 10 digits numbers.
Therefore, our number will look something like this: __ __ __ __ __ __ __ __ __ __.
There are 10 blank spaces in the above representation which needs to be filled using the digits 1 and 2.
We clearly need to repeat our digits.
So, the first place can be filled using both the digits 1 and 2. Hence, the first place has 2 ways.
Similarly, the second place can be filled using both the digits 1 and 2. Hence, second place has 2 ways.
Similarly, the third place can be filled using both the digits 1 and 2. Hence, third place has 2 ways.
Similarly, the fourth place can be filled using both the digits 1 and 2. Hence, fourth place has 2 ways.
Similarly, the fifth place can be filled using both the digits 1 and 2. Hence, fifth place has 2 ways.
Similarly, the sixth place can be filled using both the digits 1 and 2. Hence, sixth place has 2 ways.
Similarly, the seventh place can be filled using both the digits 1 and 2. Hence, seventh place has 2 ways.
Similarly, the eighth place can be filled using both the digits 1 and 2. Hence, eighth place has 2 ways.
Similarly, the ninth place can be filled using both the digits 1 and 2. Hence, ninth place has 2 ways.
Similarly, the tenth place can be filled using both the digits 1 and 2. Hence, tenth place has 2 ways.
Therefore, each of the places has 2 possibilities.
Now, multiplying these with each other.
We will get:-
$ \Rightarrow $ Number of digit we can form = \[2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = {2^{10}}\]
$\therefore $ The answer is (C) ${2^{10}}$.
Note: The students must note that we are multiplying individual place’s ways with each other to get the required number of digits. We are multiplying them because these are happening corresponding to each other.
P stands for permutation, we use it to show the shuffling of digits possible to be done in this.
C stands for combination which means the combinations we can form.
Complete step-by-step answer:
We have to form 10 digits numbers.
Therefore, our number will look something like this: __ __ __ __ __ __ __ __ __ __.
There are 10 blank spaces in the above representation which needs to be filled using the digits 1 and 2.
We clearly need to repeat our digits.
So, the first place can be filled using both the digits 1 and 2. Hence, the first place has 2 ways.
Similarly, the second place can be filled using both the digits 1 and 2. Hence, second place has 2 ways.
Similarly, the third place can be filled using both the digits 1 and 2. Hence, third place has 2 ways.
Similarly, the fourth place can be filled using both the digits 1 and 2. Hence, fourth place has 2 ways.
Similarly, the fifth place can be filled using both the digits 1 and 2. Hence, fifth place has 2 ways.
Similarly, the sixth place can be filled using both the digits 1 and 2. Hence, sixth place has 2 ways.
Similarly, the seventh place can be filled using both the digits 1 and 2. Hence, seventh place has 2 ways.
Similarly, the eighth place can be filled using both the digits 1 and 2. Hence, eighth place has 2 ways.
Similarly, the ninth place can be filled using both the digits 1 and 2. Hence, ninth place has 2 ways.
Similarly, the tenth place can be filled using both the digits 1 and 2. Hence, tenth place has 2 ways.
Therefore, each of the places has 2 possibilities.
Now, multiplying these with each other.
We will get:-
$ \Rightarrow $ Number of digit we can form = \[2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = {2^{10}}\]
$\therefore $ The answer is (C) ${2^{10}}$.
Note: The students must note that we are multiplying individual place’s ways with each other to get the required number of digits. We are multiplying them because these are happening corresponding to each other.
P stands for permutation, we use it to show the shuffling of digits possible to be done in this.
C stands for combination which means the combinations we can form.
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