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Miscellaneous Examples · Example 57

Q.A company has 5 senior officers and 7 junior commissioned officers (JCO's). A team of 4 is to be sent on a special mission. In how many ways it can be formed so that it comprises of

(i) any 4 officers?
(ii) 4 senior officers?
(iii) 2 senior and 2 junior officers (JCO's)
(iv) at least 2 senior officers
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Selecting a 4-member special-mission team from 5 senior officers and 7 JCOs (12 officers total), for four different composition requirements — a pure combinations problem since the order within the team doesn't matter.

Number of ways to choose rr objects from nn distinct objects (order irrelevant):

nCr=(nr)=n!r! (n−r)!^{n}C_r = \binom{n}{r} = \dfrac{n!}{r!\,(n-r)!}

where nn = total pool size, rr = number chosen. For a composite condition (e.g. "2 of one kind AND 2 of another"), multiply independent combination counts (multiplication principle); for "at least", sum the mutually exclusive cases.

(i) Any 4 officers (no restriction)

  1. Total officers available =5+7=12= 5 + 7 = 12.
  2. Choose any 4 of the 12: 12C4=12!4! 8!=12×11×10×94×3×2×1=1188024=495^{12}C_4 = \dfrac{12!}{4!\,8!} = \dfrac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = \dfrac{11880}{24} = 495.

(ii) Exactly 4 senior officers (and 0 junior)

  1. Choose all 4 from the 5 seniors, 0 from the 7 juniors: 5C4×7C0=5×1=5^{5}C_4 \times {}^{7}C_0 = 5 \times 1 = 5.

(iii) Exactly 2 senior and 2 junior officers

  1. Choose 2 of 5 seniors: 5C2=5×42×1=10^{5}C_2 = \dfrac{5\times4}{2\times1} = 10.
  2. Choose 2 of 7 juniors: 7C2=7×62×1=21^{7}C_2 = \dfrac{7\times6}{2\times1} = 21.
  3. Multiply (independent choices, multiplication principle): 10×21=21010 \times 21 = 210.

(iv) At least 2 senior officers (team size fixed at 4)

  1. "At least 2 senior" out of a 4-member team splits into three mutually exclusive, exhaustive cases: exactly 2, exactly 3, or exactly 4 seniors (the remaining slots filled by JCOs). …

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