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Standard XI

Mathematics

Permutation under Restriction

From 4 offi...

Question

From $4$ officers and $8$ privates, in how many ways can $6$ be chosen (1) to include exactly one officer, (2) to include at least one officer?

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Solution

Part 1:
One officer can be choosen in $^{4} C_{1}$ ways and remaining $5$ can be chosen from $8$ privates in $^{8} C_{5}$ ways
Therefore, number of ways to choose $6$ people $=^{4} C_{1} \times^{8} C_{5} = 224$

Part 2:
$6$ people can be chosen from $4$ Officers and $8$ privates in $^{12} C_{6}$ ways
$6$ people can be chosen from $8$ privates in $^{8} C_{6}$ ways
Therefore, number of ways to choose $6$ people with atleast $1$ private $=$ $^{12} C_{6} -^{8} C_{6} = 924 - 28 = 896$

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From 4 officers and 8 privates, in how many ways can 6 be chosen (1) to include exactly one officer, (2) to include at least one officer?

From $4$ officers and $8$ privates, in how many ways can $6$ be chosen (1) to include exactly one officer, (2) to include at least one officer?