1GIVEN 5 DIFERENT GREEN DYES & 4 DIFRENT BLUE dyes,3 different red dyes,no of combination of dyes taking at least 1 green dye...
A STUDENT IS ALLOWD 2 SELECT ATMOST 10 BUKS FRM COLLECTOIN OF 2N+1 BUKS .IF NO OF WAYS IN WICH HE SELECT AT LEAST 1 BUK IS 63,FIND N"
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5 Answers
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301) No. of ways to select atmost 10 books - 2n+1C0 + 2n+1C1 +2n+1C2 .... + 2n+1C10
2nd part,
No. of ways to select atleast 1 book - 22n+1 - 1
22n+1 - 1 = 63
=> 2n+1 = 7
=> n=3
2) Let the no. of green dyes be x, blue be y, and red be z.
clearly,
x≥1
y≥0
z≥0
x + y + z ≤ 12
(x'+1) + y + z ≤ 12 [ x = x' +1 , x' ≥ 0]
x' + y + z + c = 11
So required no. of ways = 14C3
262mistake in soln..... @ashish
it is 26 =64 but u have taken 7....
but on taking 6 we dont get n as a whole number....
11i think the sum will be...
a student is allowed to select at most n books from 2n+1 books...
then its solution is as
2n+1C1 + 2n+1C2 + 2n+1C3........ + 2n+1Cn = 63
we know,
2n+1C1 + 2n+1C2 + 2n+1C3........ + 2n+1Cn + 2n+1Cn+1 + 2n+1Cn+2........... + 2n+1C2n+1 = 22n+1 - 1. .................. (1)
2n+1C1 + 2n+1C2 + 2n+1C3........ + 2n+1Cn = x (say) .........................(2)
2n+1Cn+1 + 2n+1Cn+2........... + 2n+1C2n = x .....................(3) as [ nCr = nCn-r ]
and 2n+1C2n+1 = 1 ........(4)
addind equn. (2)+(3)+(4)and substituting in (1), we get,
2x + 1 = 22n+1 - 1.
x=63(given.)
126+2 = 22n+1.
22n+1 = 128 = 27.
thus, n=3. :)