The total number of natural numbers of six di | vedclass

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Question

(a) There can be two types of numbers :

$(i)$ Any one of the digits $1, 2, 3, 4$ repeats thrice and the remaining digits only once $i.e.$ of the ty

Vedclass · Accepted Answer

$1560$

Vedclass · Answer

(a) There can be two types of numbers :

$(i)$ Any one of the digits $1, 2, 3, 4$ repeats thrice and the remaining digits only once $i.e.$ of the type $1, 2, 3, 4, 4, 4$ etc.

$(ii)$ Any two of the digits $1, 2, 3, 4$ repeat twice and the remaining two only once $i.e.$ of the type $1, 2, 3, 3, 4, 4$ etc.

Now number of numbers of the $(i)$ type

$ = \frac{{6\;!}}{{3\;!}}{ 	imes ^4}{C_1} = 480$

Number of numbers of the $(ii)$ type

$ = \frac{{6\;!}}{{2\;!\;2\;!}}{ 	imes ^4}{C_2} = 1080$

Therefore the required number of 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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