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There are $6$ boxes numbered $1, 2, \dots\dots,6$. 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 :

1. $18$
2. $19$
3. $20$
4. $21$

1 comment

Why does the nPr OR nCr formula not used here?

At least one box has green ball

so green ball in 1, or 12 , 123 or 1234 or 12345 , 123456  total =6 cases

or in  2 , 23 or    234 or 2345 , 23456               total =5 cases

or in  3 , 34 or    345 or 3456                            total =4 cases

or in  4 , 45 or    456                                         total =3 cases

or in 5, 56   total 2 cases                          or  in 6 only   total 1 case

so total no of ways = 6+5+4+3+2+1=21 option D

• If one green bail in a box, then the number of ways $= 6\:(1$ box has the green ball can be any of the $6$ boxes.$)$
• If two green balls in a box, then the number of ways $= 5 \:(2$ boxes have green balls The boxes may be numbered as $12, 23, 34, 45, 56)$
• If three green balls in a box, then the number of ways $= 4\:(3$ boxes $123, 234, 345, 456)$
• If four green balls in a box, then the number of ways $= 3\:(4$ boxes $1234,2345,3456)$
• lf five green balls in a box, then the number of ways $= 2\:(5$ boxes $12345,23456 )$
• If six green balls in a box, then the number of ways $= 1\:(6$ boxes $123456)$

$\therefore$ Total number of ways $= 6 + 5 + 4 + 3 + 2 + 1 = 21.$

So, the correct answer is $(D).$

One can see how beautiful recursion is

G  -----(1)

G GG ---(2)

G GG GGG --- (3)

AND SO ON TILL

G GG … GGGGGG

F(G) = F(G-1) + 1

HERE F(1) = 1 AND WE ARE ASKED WHAT IS F(6)

SOLUTION TO THIS IS SUM OF FIRST N NATURAL NUMBERS

WHICH IS N(N+1)/2
by

can you elaborate more on your procedure ?
Later. Right now time constraints

correct ans :D  6+5+4+3+2+1=21

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1 comment

so whenever the case of consecutive numbers  we are able to use the sum of natural numbers conept