Number of six-digit numbers such that any digit that appears in the number appears at least twice, where the digits of each number are from the set `{

Correct option is: C. 18

If the sum of the digits in a number is divisible by 9, then the number itself is divisible by 9. The least and the greatest sum of five digits...

Correct Answer - B The prime digits are 2, 3, 5, 7. If we fix 2 at first place, then other 2n - 1 places are filled by all four digits....

Correct Answer - D Three-digit numbers are 100, 101, …, 999. Total number of such numbers is 900. The three-digit numbers (which have all same digits) are 111, 222, 333, …,...

Correct Answer - B Total number of cases = n(S) = 6! Now sum the given digits is 1 + 2 + 3 + 4 + 5 + 6 = 21,...

Correct Answer - D Let the probability for getting an odd number be p. Therfore the probability for getting an even number is 2p. `thereforep+2p=1orp=1/3` Sum of two numbers is even...

Correct Answer - B `(b)` Least digit used `=4` `:.` We can use `4,5,6,7,8,9`. So each of the four places can be filled in `6` ways So number of numbers is...

Correct Answer - B `(b)` `12=2^(2)*3` Then the four digits can be `1,1,6,2 to` no. of numbers `=(4!)/(2!)=12` `1,1,3,4 to` no. of numbers `=(4!)/(2!)=12` `2,2,3,1 to` no. of numbers `=(4!)/(2!)=12` Hence,...

Correct Answer - B `(b)` Required number of numbers `="(n)C_(0)(2)^(n)+"(n)C_(2)(2)^(n-2)+…` We know that `(1+x)^(n)+(x-1)^(n)=2sum_(k=0)^(n)"(n)C_(k)x^(k)` Hence, put `x=2`, get the result `2["^(n)C_(0)(2)^(n)+^(n)C_(2)(2)^(n-2)+^(n)C_(2)(2)^(n-4)+^(n)C_(6)(2)^(n-6)+...]`

Correct Answer - B `(b)` Total number of ways such that at least `3` digits will not occur in its position. `=^(5)C_(3){3!-^(3)C_(1)2!+^(3)C_(2)1!-^(3)C_(2)0!}+^(5)C_(4)[4!-^(4)C_(1)(3!)+^(4)C_(2)(2!)-^(4)C_(3)(1!)+^(4)C_(4)(0!)}+^(5)C_(5){5!-^(5)C_(1)4!+^(5)C_(2)3!-^(5)C_(3)2!+^(5)C_(4)1!-^(5)C_(5)(0!)}` `=10(2)+5(9)+(44)` `=20+45+44=109`