Question

Show the total number of natural numbers of six digits that can be made with digits. 1,2,3,4, if all numbers are to appear in the same number at least once is 1560.

Answer 1

Hint: In this question we use the theory of permutation and combination. We have to choose numbers of 6 digits out of given 4 digits using all the four. This is possible for some of the digits to repeat to make 6 digits. Only one number out of four will appear three times. This can be done in${}^{\text{4}}{{\text{C}}_{\text{1}}}$ =4 ways.

Complete Step-by-Step solution:
Out of the six digits, four digits are fixed.
In the remaining two digits, both the digits could be same or both the digits could be different as well
Case 1 - When both the digits are the same.
Number of six-digit numbers =${}^{\text{4}}{{\text{C}}_{\text{1}}}{\text{} \times }\dfrac{{{\text{6!}}}}{{{\text{3!}}}}$
(Both the digits could be any of the four digits, therefore ${}^{\text{4}}{{\text{C}}_{\text{1}}}$)
Case 2 - When both the digits are different.
(The two digits could be any two of the four digits, therefore ${}^{\text{4}}{{\text{C}}_2}$)
Therefore, the number of six-digit numbers in this case = ${}^{\text{4}}{{\text{C}}_2}$ × $\dfrac{{6!}}{{2!{\text{ }} \times {\text{ }}2!}}$
Thus, the required number of 6 digit numbers = ${}^{\text{4}}{{\text{C}}_{\text{1}}}{\text{} \times }\dfrac{{{\text{6!}}}}{{{\text{3!}}}}$+ ${}^{\text{4}}{{\text{C}}_2}$ × $\dfrac{{6!}}{{2!{\text{ }} \times {\text{ }}2!}}$
=6! (23+32)
= 6! × 136
= 1560
Thus, the required number is 1560.
Therefore, we prove that the total number of natural numbers of six digits that can be made with digits 1,2,3,4, if all numbers are to appear in the same number at least once is 1560.

Note: In this question, you need to know first, basics of permutation and combination theory and also to prevent undesirable errors, we need to take care of the calculation part also. & note that Number of 6-digit number that can be formed without 1 is \[{\text{}3 \times 3 \times 3 \times 3 \times 3 \times 3}\] = 729.This is because each place now can be filled with only 3 remaining numbers.