there are 6 boxes numbered 1 , 2 , … , . 6 . each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered . the total number of ways in which this can be done is
list down possibilities : from only 1 box all the way to all 6 if only one of the boxes has a green ball , it can be any of the 6 boxes . so , we have 6 possibilities . if two of the boxes have green balls and then there are 5 consecutive sets of 2 boxes . 12 , 23 , 34 , 45 , 56 . if 3 of the boxes have green balls , there are 4 possibilities : 123 , 234 , 345 , 456 . if 4 boxes have green balls , there are 3 possibilities : 1234 , 2345 , 3456 . if 5 boxes have green balls , there are 2 possibilities : 12345 , 23456 . if all 6 boxes have green balls , there is just 1 possibility . total number of possibilities = 6 + 5 + 4 + 3 + 2 + 1 = 21 . ans : b
divide(multiply(6, add(6, 1)), 2)