Digit at hundred’s place is always greater than the digit at ten’s as well as unit’s place.
Possible numbers for hundred’s place = 9, 7 and 5
If we select ‘9’ at hundred’s place, other places can have four possibilities (2, 3, 5 and 7)
i. e. 4C2 = 6
If we select ‘7’ at hundred’s place, other places can have three possibilities (2, 3 and 5)
i. e. 3C2 = 3
If we select ‘5’ at hundred’s place, other places can have two possibilities (2 and 3)
i. e. 2C2 = 1
Total possibilities = 6 + 3 + 1 =10 ways.
Initially -> 100 After first month -> 100 * 3 After second month -> 100 * 3 * 3 After third month -> 100 * 3 * 3 * 3 After fourth month -> 100 * 3 * 3 * 3 * 3 => 100 * 3^4
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16,800/42=400 400=(20)^2 Ans-42
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The balls are numbered as 1,2,3,4,5,6,7,8,9 and 10. If 1 numbered ball picks up by Anuj then Anisha will picks up 2 or 3 or 4 or ..... or 10 = 9 ways. like this, 2 numbered ........................... = 8 ways. 3 numbered ............................ = 7 ways. . . . 9 numbered................................ = 1 way So, number of ways that Anuj picks a ball numbered less than that picked by Anisha = 9+8+7+6+5+4+5+4+3+2+1 = 45 ways Amit picked a ball from remaining 8 balls = 8c1 = 8 ways one ball lesser from 3, that is 3c1 number of possible ways = (45*8)/3= 120 Total number of possible ways = 10c1 * 10c1 *10c1 = 1000 The probability = 120/1000 = 3/25
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Using nCr p^r q^n-r we get 7c2(0.40)^2*(0.60)^5
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