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Comments (98)
U can make the solution simpler ..
try using summations.
SOLUTION: ( just try to select text.)
2x+y+z = 20 ..................(1)
condition on x ::
x is max when y and z are minimum...
2max(x) = 20 - min(y) - min(z)
max(x) = (20 - 0 - 0) / 2 = 10
Let x = k...where 0<=k<=10
Put x =k in (1) to get
y+z = 20 - 2k ...................(2)
NUmber of non negative integral solutions of (2) = 20 - 2k + 1
= 21 - 2k
As k varies from 0 to 10, the total number of non negative solutions of (1)
= summ{k =0 to k = 10} ( 21 - 2k) = summ{k=0 to k=10} (21) - 2.summ{k=0 to k=10}(k)
= 231 - 110 = 121
( using ..summ(n) = n(n+1) / 2 )
Hence total number of non negative integral solutions od (1) = 121.
Solution ends here.
ans is 50 ...Want solution ? or Let others try..?
Also number your questions so that we have a proper record.
this one is a nut-cracker.
4. There are 12 seats in the first row of a theater of which 4 are to be occupied. Finf the number of ways of arranging 4 persons so that:
a) No two persons sit side by side.
b) There hould be atleast 2 empty seats between any two person.
c) Each person has exactly one neighbour.
solving b
x1 A x2 B x3 C x4 D x5....where xi are no of empty seats......A,B,C,D are persons........
x1+x2+x3+x4=8,x1>=0,x2,x3,x4>=2,x5>=0.........now using multinomial thm.
x1+x2+2+x3+2+x4+2+x5=8........
no. of non negative integral solns. of x1+x2+x3+x4+x5=2........=2+5-1C5-1=15.is dat tru.
Answer to the first part is 9P4 .
Solution ( try to select text.)...
a) We have to select 4 seats for 4 persons so that no 2 persons are together. It means that there should be atleast one empty seat vacant between any two persons.
To place 4 persons, we have to select 4 seats between the remaining 8 empty seatsso that all the persons should be separated.
Between 8 empty seats 9 seats are available for 4 persons to sit.
Select 4 seats in 9C4 ways.
And arrange 4 persons in 4! ways ..
Hence total no. of ways = 9C4.4! = 9P4.
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Okay hotshot ..lets dedicate this post to the PNCians ( not to be confused with VMCians) ..
Questions coming straight from my " ewww! what's that ? "-list.
1.Show that the number of ways of selecting n things out of 3n things of which n are of one kind and alike and n are of a second kind and alike and the rest are unalike is (n+2).2^(n-1)