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Further Mathematics Paper 2, Nov/Dec. 2008  
Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Main
General Comments
Weakness/Remedies
Strength
Question 17

(a)(i)  In how many ways can eight boys be seated in a row?
(ii)  If two of the boys in (a) (i) cannot sit together, in how many ways can they be selected?    
      
(b)  A delegation of a labour union consist of 7 men and 3 women.  If 4 of them are selected at random to give a talk, what is the probability of selecting:

  • 3 men and 1 woman
  • at least 2 women?
_____________________________________________________________________________________________________
Observation

The part (a) of the questions was well attempted by majority of the candidates.  They were able to recall that the number of ways in which 2 boys must not sit together is equal to no of ways of sitting 8 people minus no of ways if 2 boys must sit together which is
8! –(2 x 7!)  = 30240 ways.

In the part (b), they were able to obtain the probability of selecting 3 men and 1 woman as      7C 3   x   3C1     =   ½   and the probability of selecting at least 2 women as                                                                                10C4

        7C2  x  3C2  +  7C1  x  3C2         =  1/3  or  0.333
                           10C4

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