Restricted selection / arrangement:
(a) The number of types in which r objects may be chosen from n different types of objects if k particular objects are
(a) always added = n-k Cr-k
(b) never added = n-k Cr
(b) The number of arrangements of n distinct objects taken r at a time so that k spacific objects are
(a) always added = n-k Cr-k .r!
(b) never added = n-k Cr .r!
Example: A delegation of four students is to be chosen from a total of 12 students. In how many types may the delegation be chosen
(a) If all the students are identically willing.
(b) If two particular students have to be added in the delegation.
(c) If two particular students do not need to be together in the delegation.
(d) If two particular students want to be included together only.
(e) If two particular students did not want to be together and two other specific student want to be together only in the delegation.
Solution: (a) Creation of delegation denotes selection of 4 out of 12. Therefore the number of types = 12C4 = 495.
(b) If two specific students are already chosen. Here we want to choose only 2 out of the outstanding 10. Therefore the number of types = 10C2 = 45.
(c) The number of types in which both are chosen = 45. Therefore the number of types in which the two are not added together = 495 - 45 = 450.
(d) There are two probable cases
(i) Either both are chosen. In that case, the number of types in which the selection may be prepared = 45.
(ii) Or both are not chosen. In that case all the four students are chosen from the remaining ten students.
That may be completed in 10C4 = 210 types.
Therefore the total number of types of selection = 45 + 210 = 255.
(e) We consider that students A and B want to be chosen together and students C and D do not want to be together. Now there are subsequent 6 types.
(i) (A, B, C) selected, (D) not selected
(ii) (A, B, D) selected (C) not selected
(iii) (A, B) selected (C, D) not selected
(iv) (C) selected (A, B, D) not selected
(v) (D) selected (A, B, C) not selected
(vi) A, B, C, D not selected
For (i) the number of ways of selection = 8C1 = 8
For (ii) the number of ways of selection = 8C1 = 8
For (iii) the number of ways of selection = 8C2 = 28
For (iv) the number of ways of selection = 8C3 = 56
For (v) the number of ways of selection = 8C3 = 56
For (vi) the number of ways of selection = 8C4 = 70
Therefore total number of types = 8 + 8 + 28 + 56 + 56 + 70 = 226
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