01. a number of 4 different digits is formed by using the digits 1, 2, 3, 4, 5, 6,7, 8 in all possible
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1 01. a number of 4 different digits is formed by using the digits 1, 2, 3, 4, 5, 6,7, 8 in all possible. Find how many numbers are greater than 3000 thousand place can be filled by any one of the digits 3, 4, 5, 6, 7, 8 in 6 P1 Having done that the remaining 3 places can be filled by any 3 of the remaining 7 digits in 7 P3 Total numbers formed = 6 P1 x 7 P3 = 6 x 7 x 6 x 5 = a number of 4 different digits is to be formed by using the digits 1, 2, 3, 4, 5, 6, 7, 8, 9. Find how many of them are a) greater than 4000 thousand place can be filled by any one of the digits 4, 5, 6, 7, 8, 9 in 6 P1 Having done that the remaining 3 places can be filled by any 3 of the remaining 8 digits in 8 P3 Total numbers formed = 6 P1 x 8 P3 = 6 x 8 x 7 x 6 = 2016 b) divisible by 2 unit place can be filled by any one of the digits 2, 4, 6, 8 in 4 P1 Having done that the remaining 3 places can be filled by any 3 of the remaining 8 digits in 8 P3 Total numbers formed = 4 P1 x 8 P3 = 4 x 8 x 7 x 6 = 1344 c) divisible by 5 unit place can be filled by digit 5 in 1 way Having done that the remaining 3 places can be filled by any 3 of the remaining 8 digits in 8 P3 Total numbers formed = 1 x 8 P3 = 8 x 7 x 6 =
2 03. How many 5 different digit numbers can be formed with digits 2, 3, 5, 7, 9 which are a) greater than thousand place can be filled by any one of the digits 3, 5, 7, 9 in 4 P1 Having done that the remaining 4 places can be filled by remaining 4 digits in 4 P4 = 4! Total numbers formed = 4 P1 x 4! = 4 x 24 = 96 b) less than thousand place can be filled by any one of the digits 2, 3, 5 in 3 P1 Having done that the remaining 4 places can be filled by remaining 4 digits in 4 P4 = 4! Total numbers formed = 3 P1 x 4! = 3 x 24 = 72 b) between & thousand place can be filled by any one of the digits 3, 5, 7 in 3 P1 Having done that the remaining 4 places can be filled by remaining 4 digits in 4 P4 = 4! Total numbers formed = 3 P1 x 4! = 3 x 24 = how many different digit numbers can be formed between 100 and 1000 using 0, 1, 3, 5 and 7 which is not divisible by 5 unit place can be filled by any one of digits 1, 3 & 7 in 3 P1 Hundreds place can be filled by any one the remaining 3 digits ( 0 excluded) in 3 P1 tens place can then be filled by any one of the remaining 3 digits in 3 P1 Total numbers formed = 3 P1 x 3 P1 x 3 P1 = 3 x 3 x 3 = How many different digit numbers are formed between 7000 and 8000 using 0, 1, 3, 5, 7 and 9 which are divisible by 5 thousand place can be filled by digit 7 in 1 way units place can be filled by any on e of the digits 0, 5 in 2 P1 remaining 2 places can be filled by any 2 of the remaining 4 digits in 4 P2 Total numbers formed = 1 x 2 P1 x 4 P2 = 1 x 2 x 4 x 3 =
3 06. how many even numbers of four digits can be formed using digits 0, 1, 2, 3, 4, 5 and 6, no digit being used more than once Case 1 : Numbers ending with 0 Unit place can be filled by digit 0 in one way Having done that the remaining 3 places can be filled by any 3 of remaining 6 digits in 6 P3 Numbers formed = 1 x 6 P3 = 6 x 5 x 4 = 120 Case 2 : Numbers ending with 2, 4, 6 Unit place can be filled by any one of digits 2, 4, 6 in 3 P1 Thousand place can be filled by any one the remaining 5 digits ( 0 excluded) in 5 P1 Having done that the remaining 2 places can be filled by any 2 of remaining 5 digits in 5 P2 By fundamental princi ple of Multipliation, Numbers formed = 3 P1 x 5 P1 x 5 P2 = 3 x 5 x 5 x 4 = 300 By fundamental principle of ADDITION Total numbers formed = = how many 5 different digit numbers can be formed with digits 0, 1, 3, 5, 6, 8 and 9 divisible by 5 Case 1 : Numbers ending with 0 Unit place can be filled by digit 0 in one way Having done that the remaining 4 places can be filled by any 3 of remaining 6 digits in 6 P4 Numbers formed = 1 x 6 P4 = 6 x 5 x 4 x 3 = 360 Case 2 : Numbers ending with 5 Unit place can be filled by digit 5 in 1 ten Thousand place can be filled by any one the remaining 5 digits ( 0 excluded) in 5 P1 Having done that the remaining 3 places can be filled by any 3 of remaining 5 digits in 5 P3-3 -
4 Numbers formed = 1 x 5 P1 x 5 P3 = 1 x 5 x 5 x 4 x 3 = 300 By fundamental principle of ADDITION Total numbers formed = =
5 NOV 2015 Find the coordinates of the orthocenter of a triangle whose vertices are ( 2,3), (6, 1), (4,3) y 3 = 2 (x 4) y 3 = 2x 8 2x y = 5 ALTITUDE AD A ( 2,3) ORTHOCENTER H x 2y = 8 2x y = 5 x 2 B(6, 1) D C(4,3) mbc = y2 y1 x2 x1 = = 4 = AD m = 1 (AD BC), A( 2,3) 2 x 2y = 8 4x 2y = 10 3x = 18 x = 6 subs in (1) y = 7 H (6,7) y y1 = m(x x1) y 3 = 1 (x + 2) 2 2y 6 = x + 2 x 2y = 8 ALTITUDE CE E A( 2,3) B(6, 1) D C(4,3) mab = y2 y1 x2 x1 = = 4 = CE m = 2 (CE AB), C(4,3) y y1 = m(x x1) - 5 -
6 02. The points A(2,3), B(4, 1) and C( 1,2) are the vertices of ABC. Find the length of the perpendicular from C on AB and hence find then area of ABC BASE (AB) = (x1 x2) 2 + (y1 y2) 2 = (2 4) 2 + (3 + 1) 2 Equation of AB = C ( 1,2) = 20 = 2 5 A(2,3) D B(4, 1) AREA OF ( ABC) y y1 = y2 y1 (x x1) x2 x1 y 3 = 1 3 ( x 2) 4 2 y 3 = 2(x 2) = 1 x BASE x HEIGHT 2 = 1 x 2 5 x = 7 sq. units y 3 = 2x + 4 2x + y 7 = 0 the length of the perpendicular from C on AB Height (H) C ( 1,2) 2x + y 7 = 0 A(2,3) D B(4, 1) H = 2( 1) = = 7 5 =
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