From 4 officers and 8 jawans in how many ways can 6 be chosen (i) to include exactly one officer (ii) to include at least one officer?

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Solution

(i) From 4 officers and 8 jawans, 6 need to be chosen. Out of them, 1 is an officer.
Required number of ways = $^{4} C_{1} \times^{8} C_{5} = 4 \times \frac{8!}{5! 3!} = 4 \times \frac{8 \times 7 \times 6 \times 5!}{5! \times 6} = 224$

(ii) From 4 officers and 8 jawans, 6 need to be chosen and at least one of them is an officer.
Required number of ways = Total number of ways $-$ Number of ways in which no officer is selected
$=^{12} C_{6} -^{8} C_{6} = \frac{12!}{6! 6!} - \frac{8!}{6! 2!} = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1} - \frac{8 \times 7}{2} = 924 - 28 = 896$

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