668 is correct for sweets questionIsaac attempts all six questions on an Olympiad paper in order. Each question is marked on a scale from 0 to 10. He never scores more in a later question than in any earlier question. How many dierent possible sequences of six marks can he achieve?
No. of 10 pointers = m10 No. of 9 pointers = m9 . . . . No. of 0 pointers = m0
m0 + m1 + m2 + ... + m10 = 6 -> C(16,10)
Given any of these set of marks, they can be arranged in only 1 order.
The number of ways in which all the integers from 1 to 36 (both inclusive) can be arranged such that no two multiples of 6 are adjacent is expressed as m! —n Pr where m, n, r are distinct positive integers. What is the sum m + n + r ?OPTIONS1) 66 2) 67 3) 68 4) None
Product of the fourth term and the fifth term of an arithmetic progression is 456.Division of the ninth term by the fourth term gives quotient as 11 and remainder as 10.Find the first term.
A can build a wall in the same time in which B and C together can. If A and B together do it in 25 days and C alone can do it in 35 days. In what time will B alone do it?17510090none of the above
A can build a wall in the same time in which B and C together can. If A and B together do it in 25 days and C alone can do it in 35 days. In what time will B alone do it?17510090none of the above
668 is correct for sweets questionIsaac attempts all six questions on an Olympiad paper in order. Each question is marked on a scale from 0 to 10. He never scores more in a later question than in any earlier question. How many dierent possible sequences of six marks can he achieve?
Can only think of the horrible long method :banghead:
A can build a wall in the same time in which B and C together can. If A and B together do it in 25 days and C alone can do it in 35 days. In what time will B alone do it?17510090none of the above
175 a b c be no.of units per day by A B C respectively 25(a+b)=35c
a=b+c
25(2b+c)=35c b=c/5 time taken by B=5 time of C =175 Days
The number of ways in which all the integers from 1 to 36 (both inclusive) can be arranged such that no two multiples of 6 are adjacent is expressed as m! —n Pr where m, n, r are distinct positive integers. What is the sum m + n + r ?OPTIONS1) 66 2) 67 3) 68 4) None
there are total 6 multiples of 6
1st arrange remaining 30 in 30! ways, then chose 6 places out of 31 and can arrange them in 6! ways
No. of 10 pointers = m10No. of 9 pointers = m9....No. of 0 pointers = m0m0 + m1 + m2 + ... + m10 = 6 -> C(16,10)Given these set of marks, they can be arranged in only 1 order.So total ways = C(16,10)
My approach - Divide 10 marks into 7 groups (before 1st paper, between 1st and 2nd and so on). Groups can be empty. Then starting from 10, if we subtract the values obtained in order, it would give us 6 marks (the seventh subtraction would give 0). Hence (10+6)C6 ways.
4²² - C(4, 1)*3²² + C(4, 2)*2²² - 4 is correct for the last questionFor detailed explanations read about Principle of Inclusion - Exclusion (derangements is one of the example of this principle)onlinemathcircle.com/wp-conten...www.math.cornell.edu/~putnam/i...In a school, there are 4 more students in Class A than in Class B, and 4 more students in Class B than in Class C. As a reward for good behaviour a teacher decides to give out some sweets. Every student in class C got 5 more sweets than every student in Class B, and every student in Class B got 3 more sweets than every student in Class A. As it turns out, Class A got 10 more sweets than Class B, and Class B got 14 more sweets than Class C. How many sweets were given out in all?
A can build a wall in the same time in which B and C together can. If A and B together do it in 25 days and C alone can do it in 35 days. In what time will B alone do it?17510090none of the above
C(16, 6) or C(16, 10) is correct for the marks question (as explained by @Angadbir and @scrabbler )
Let S be the set of proper three digit positive integers less than 1000. What is the minimum number of integers that must be chosen from S to ensure that there are at least two integers which have at least one digit in common?
A can build a wall in the same time in which B and C together can. If A and B together do it in 25 days and C alone can do it in 35 days. In what time will B alone do it?17510090none of the above
1/a +1/b = 1/25; 1/c = 1/35
1/a - 1/b = 1/c = 1/35; solving for a and b, we get; a = 175/6 and b = 175
Let S be the set of proper three digit positive integers less than 1000. What is the minimum number of integers that must be chosen from S to ensure that there are at least two integers which have at least one digit in common?
10? We could get 9 without repeats as 111, 222, 333...
(if the numbers have distinct digits then I guess 4) regards scrabbler
668 is correct for sweets questionIsaac attempts all six questions on an Olympiad paper in order. Each question is marked on a scale from 0 to 10. He never scores more in a later question than in any earlier question. How many dierent possible sequences of six marks can he achieve?
Product of the fourth term and the fifth term of an arithmetic progression is 456.Division of the ninth term by the fourth term gives quotient as 11 and remainder as 10.Find the first term.Any shortcut methods plz
