JEE MAIN - Mathematics (2021 - 26th February Evening Shift - No. 14)
A seven digit number is formed using digits 3, 3, 4, 4, 4, 5, 5. The probability, that number so formed is divisible by 2, is :
$${1 \over 7}$$
$${4 \over 7}$$
$${6 \over 7}$$
$${3 \over 7}$$
Explanation
Digits = 3, 3, 4, 4, 4, 5, 5
Total 7 digit numbers = $${{7!} \over {2!2!3!}}$$
Number of 7 digit number divisible by 2 $$ \Rightarrow $$ last digit = 4
_26th_February_Evening_Shift_en_14_2.png)
Now 7 digit numbers which are divisible by 2 = $${{6!} \over {2!2!2!}}$$
Required probability = $${{{{6!} \over {2!2!2!}}} \over {{{7!} \over {3!2!2!}}}} = {3 \over 7}$$
Total 7 digit numbers = $${{7!} \over {2!2!3!}}$$
Number of 7 digit number divisible by 2 $$ \Rightarrow $$ last digit = 4
Now 7 digit numbers which are divisible by 2 = $${{6!} \over {2!2!2!}}$$
Required probability = $${{{{6!} \over {2!2!2!}}} \over {{{7!} \over {3!2!2!}}}} = {3 \over 7}$$
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