The total number of natural numbers of six di | vedclass

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(A) To form a $6$-digit number using digits ${1, 2, 3, 4}$ such that each digit appears at least once,we have two possible cases for the distribution

Vedclass · Accepted Answer

$1560$

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(A) To form a $6$-digit number using digits ${1, 2, 3, 4}$ such that each digit appears at least once,we have two possible cases for the distribution of digits:Case $(i)$: One digit appears $3$ times,and the other three digits appear $1$ time each (e.g.,$1, 1, 1, 2, 3, 4$).Number of ways to choose the digit that repeats $3$ times is $^4C_1 = 4$.The number of arrangements for each selection is $\frac{6!}{3!1!1!1!} = \frac{720}{6} = 120$.Total for Case $(i) = 4 	imes 120 = 480$.Case $(ii)$: Two digits appear $2$ times each,and the other two digits appear $1$ time each (e.g.,$1, 1, 2, 2, 3, 4$).Number of ways to choose the two digits that repeat $2$ times is $^4C_2 = 6$.The number of arrangements for each selection is $\frac{6!}{2!2!1!1!} = \frac{720}{4} = 180$.Total for Case $(ii) = 6 	imes 180 = 1080$.Total number of such $6$-digit numbers $= 480 + 1080 = 1560$.

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

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