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Question 9

Mathematics Permutation and Combination Hard

The number of ways in which 5 days can be chosen in each of the 12 months of a non-leap year, is:

(A) (<sup>30</sup>C<sub>5</sub>)<sup>4</sup> × (<sup>31</sup>C<sub>5</sub>)<sup>7</sup> × (<sup>28</sup>C<sub>5</sub>)

(B) (<span style="position: relative; line-height: 0; vertical-align: baseline; top: -0.5em;font-size:10.5px;">30</span>C<span style="position: relative; line-height: 0; vertical-align: baseline; bottom: -0.25em;font-size:10.5px;">5</span>)<span style="position: relative; line-height: 0; vertical-align: baseline; top: -0.5em;font-size:10.5px;">6</span> × (<span style="position: relative; line-height: 0; vertical-align: baseline; top: -0.5em;font-size:10.5px;">28</span>C<span style="position: relative; line-height: 0; vertical-align: baseline; bottom: -0.25em;font-size:10.5px;">5</span>)<sup>6</sup>

(C) (<span style="position: relative; line-height: 0; vertical-align: baseline; top: -0.5em;font-size:10.5px;">30</span>C<span style="position: relative; line-height: 0; vertical-align: baseline; bottom: -0.25em;font-size:10.5px;">5</span>)<span style="position: relative; line-height: 0; vertical-align: baseline; top: -0.5em;font-size:10.5px;">7</span> × (<span style="position: relative; line-height: 0; vertical-align: baseline; top: -0.5em;font-size:10.5px;">31</span>C<span style="position: relative; line-height: 0; vertical-align: baseline; bottom: -0.25em;font-size:10.5px;">5</span>)<span style="position: relative; line-height: 0; vertical-align: baseline; top: -0.5em;font-size:10.5px;">4</span> × (<span style="position: relative; line-height: 0; vertical-align: baseline; top: -0.5em;font-size:10.5px;">28</span>C<span style="position: relative; line-height: 0; vertical-align: baseline; bottom: -0.25em;font-size:10.5px;">5</span>)

(D) (<span style="position: relative; line-height: 0; vertical-align: baseline; top: -0.5em;font-size:10.5px;">30</span>C<span style="position: relative; line-height: 0; vertical-align: baseline; bottom: -0.25em;font-size:10.5px;">5</span>)<span style="position: relative; line-height: 0; vertical-align: baseline; top: -0.5em;font-size:10.5px;">5</span> × (<span style="position: relative; line-height: 0; vertical-align: baseline; top: -0.5em;font-size:10.5px;">31</span>C<span style="position: relative; line-height: 0; vertical-align: baseline; bottom: -0.25em;font-size:10.5px;">5</span>)<span style="position: relative; line-height: 0; vertical-align: baseline; top: -0.5em;font-size:10.5px;">6</span> × (<span style="position: relative; line-height: 0; vertical-align: baseline; top: -0.5em;font-size:10.5px;">28</span>C<span style="position: relative; line-height: 0; vertical-align: baseline; bottom: -0.25em;font-size:10.5px;">5</span>)

✓ Correct Answer: Option A Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

1. Distribution of Days in a Non-Leap Year: \[ \text{7 months have 31 days.} \] \[ \text{4 months have 30 days.} \] \[ \text{1 month has 28 days.} \] 2. Calculate the Number of Ways to Choose 5 Days: \[ \text{For each of the 7 months with 31 days, the number of ways to choose 5 days is } \binom{31}{5}. \] \[ \text{For each of the 4 months with 30 days, the number of ways to choose 5 days is } \binom{30}{5}. \] \[ \text{For the 1 month with 28 days, the number of ways to choose 5 days is } \binom{28}{5}. \] 3. Combine the Results: \[ \text{The total number of ways is:} \] \[ \left( \binom{31}{5} \right)^7 \times \left( \binom{30}{5} \right)^4 \times \binom{28}{5} \] Therefore, the correct option is: \[ \boxed{\left( \binom{31}{5} \right)^7 \times \left( \binom{30}{5} \right)^4 \times \binom{28}{5}} \]

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