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How many permutations of the 10 digits either begin with the 3 digits 987, contain the digits 45 in the fifth and sixth positions, or end with the 3 digits 123?
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5158 ?

numbers starting with 978 => 987_ _ _ _ _ _ _ => 7!
numbers having 45 in 5th and 6th position => _ _ _ _ 45_ _ _ _ => 8!
numbers ending with 123=7!

numbers starting with 978 and having 45 in 5th and 6th position=98745_ _ _ _ _ =>5!
numbers starting with 978 and ending with 123 => 978 _ _ _ _ 123 =>4!
numbers having 45 in 5th and 6th position and ending with 123 => _ _ _ 45 _ _ 123 =>5!

numbers starting with 978, having 45 in 5th and 6th position and ending with 123 => 98745_ _ 123 =>2!

7!+8!+7!-5!-4!-5!+2!

=50,138
by Active (5.1k points)
+1 vote
$($begin with the 3 digits 987, contain the digits 45 in the fifth and sixth positions$)$ $or$ $($end with the 3 digits 123 $)$

let the $10$ digits be represented as _ _ _ _ _ _ _ _ _ _

987_ 45_ _ _ _  = $5!$

_ _ _ _ _ _ _ 123 = $7!$

987_45_ 123   = $2!$

$N(A\cup B)$

= $N(A)+N(B)-N(A\cap C)$

= $5! +7! -2!$

=$120 +5040 - 2$

=$5158$
by Boss (21.8k points)
edited by
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@Satbir digits can be repeated here right?

+1

I have not considered repetition of digits. coz it says permutations of 10 digits so i felt that we have to use all the digits in each case.

if repetition is allowed

Then we can fill every blank space in 10 ways

So it will become $10^{5} + 10^{7} - 10^{2}$  right ?

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yes correct
+1
i think u interpreted it wrongly
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So what would have been written in the question for the way which i have solved ? and how did you came to know that there are three cases not two ? @aditi19

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nothing could be written. The question is itself slightly ambiguous