
The number of numbers greater than 50,000 that can be formed by using digits 3, 5, 6, 6, 7, 9 is:
A) 47
B) 48
C) 50
D) None of these
Answer
495.6k+ views
Hint:
Here the digits given are 6 and we have to make numbers greater than 50000. So except 3 all other numbers can take place of ten thousandth value. So, place them and use permutations or combinations. Also remember that 6 is the digit repeated in the given numbers.
Complete step by step solution:
Here they have given that the number should be greater than 50,000. So the numbers formed will contain 5 digits. And 6 is the digit repeated.
So, combinations formed by digit 5 \[ = \dfrac{{5!}}{{2!}}\]
\[
\Rightarrow \dfrac{{120}}{2} \\
\Rightarrow 60 \\
\]
So ,combinations formed by the digits are 60.but we also included the numbers formed by 3 on ten thousandth place (extreme left) .But those numbers are less than 50000. So we need to subtract them from total possible combinations.
So 3 will be at extreme left and remaining 4 digits will form the combination (remember 6 is repeated)
\[
\Rightarrow \dfrac{{4!}}{{2!}} \\
\Rightarrow \dfrac{{24}}{2} \\
\Rightarrow 12 \\
\]
Now we will subtract these from total combinations.
\[ \Rightarrow 60 - 12 = 48\]
So option B is correct.
Note:
Remember that given digits are 6 but we have to form a 5 digit number. Numbers formed by 3 are not considered because they are not greater than 50000. Also note that 6 is a repeated digit so the combination should be divided by $2!$
Combination is used in conditions of picking balls , picking a card , selecting a committee, forming a team, forming numbers.
Here the digits given are 6 and we have to make numbers greater than 50000. So except 3 all other numbers can take place of ten thousandth value. So, place them and use permutations or combinations. Also remember that 6 is the digit repeated in the given numbers.
Complete step by step solution:
Here they have given that the number should be greater than 50,000. So the numbers formed will contain 5 digits. And 6 is the digit repeated.
So, combinations formed by digit 5 \[ = \dfrac{{5!}}{{2!}}\]
\[
\Rightarrow \dfrac{{120}}{2} \\
\Rightarrow 60 \\
\]
So ,combinations formed by the digits are 60.but we also included the numbers formed by 3 on ten thousandth place (extreme left) .But those numbers are less than 50000. So we need to subtract them from total possible combinations.
So 3 will be at extreme left and remaining 4 digits will form the combination (remember 6 is repeated)
\[
\Rightarrow \dfrac{{4!}}{{2!}} \\
\Rightarrow \dfrac{{24}}{2} \\
\Rightarrow 12 \\
\]
Now we will subtract these from total combinations.
\[ \Rightarrow 60 - 12 = 48\]
So option B is correct.
Note:
Remember that given digits are 6 but we have to form a 5 digit number. Numbers formed by 3 are not considered because they are not greater than 50000. Also note that 6 is a repeated digit so the combination should be divided by $2!$
Combination is used in conditions of picking balls , picking a card , selecting a committee, forming a team, forming numbers.
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