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# Find the number of numbers greater than ${10^6}$ that can be formed using the digits of the number 2334203, if all the digits of the given number must be used.

Last updated date: 13th Jun 2024
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Hint: The formula of permutations with repetition, that is, $\dfrac{{{}^n{P_r}}}{{{x_1}!{x_2}!}}$, where $n$ is the number of digits, $r$ is the number of required digits and ${x_i}$’s are the number of times a letter is repeated. Apply this formula, and then use the given conditions to find the required value.

Complete step-by-step solution:
Given that the number is 2334203.
Since we know that the number 3 is repeated three times and 2 is repeated two times in the above number.

So, we use the formula of permutations with repetition, that is, $\dfrac{{{}^n{P_r}}}{{{x_1}!{x_2}!}}$, where $n$ is the number of digits, $r$ is the number of required digits and ${x_i}$’s are the number of times a letter is repeated.

We will now find the total number of permutations of the digits of the given number from the above formula of permutations with repetition.

$\dfrac{{{}^7{P_7}}}{{2!3!}} = \dfrac{{7!}}{{2!3!}}$

Calculating the factorials of the above fraction, we get

$\dfrac{{7 \times 6 \times 5 \times 4 \times 3!}}{{2 \times 1 \times 3!}} = \dfrac{{7 \times 6 \times 5 \times 4}}{2} \\ = 7 \times 6 \times 5 \times 2 \\ = 420 \\$

Since we know that when 0 will take the first position, the number becomes a 6 digit number which is not greater than ${10^6}$.

We will now find the number of permutations of 6 digit numbers from the above formula of permutations with repetition.

$\dfrac{{{}^6{P_6}}}{{2!3!}} = \dfrac{{6!}}{{2!3!}}$

Calculating the factorials of the above fraction, we get

$\dfrac{{6 \times 5 \times 4 \times 3!}}{{2 \times 1 \times 3!}} = \dfrac{{6 \times 5 \times 4}}{2} \\ = 6 \times 5 \times 2 \\ = 60 \\$

Now, we will find the number of numbers, which are greater than ${10^6}$ from the number 2334203.

Subtracting the number of permutations of 6 digit numbers from total number of permutations, we get

$420 - 60 = 360$
Thus, the number of numbers greater than ${10^6}$ that can be formed using all the digits of the number 2334203 is 360.

Note: In solving these types of questions, you should be familiar with the formula to find the permutations with repetition. Then use the given conditions and values given in the question, and substitute in the formula for permutations, to find the values. Also, we are supposed to write the values properly to avoid any miscalculation.