The Solution:
Given:
7 students from the Junior class.
6 students from the Senior class.
4 new members are to be chosen.
Required:
Find the number of ways 4 new members can be chosen if 2 or fewer must be from the senior class.
So, the possibilities are:
![\begin{gathered} (^6C_0\cdot^7C_4)\text{ or }(^6C_1\cdot^7C_3)\text{ or }(^6C_2\cdot^7C_2) \\ \\ (^6C_0\cdot^7C_4)+(^6C_1\cdot^7C_3)+(^6C_2\cdot^7C_2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/jnqjyffxfhaekz73y2k64op9uqhqhtel27.png)
By formula, Combination is
So,
![\lbrack(6!)/((6-0)!0!)*(7!)/((7-4)!4!)\rbrack+\lbrack(6!)/((6-1)!1!)*(7!)/((7-3)!3!)\rbrack+\lbrack(6!)/((6-2)!2!)*(7!)/((7-2)!2!)\rbrack](https://img.qammunity.org/2023/formulas/mathematics/college/pgcg87fi3bwsaui8ve31hnflyj1ulyfsnl.png)
![=(1*35)+(6*35)+(15*21)=35+210+315=560\text{ ways}](https://img.qammunity.org/2023/formulas/mathematics/college/wgazm7ie6202vns8m3m296ujlun5k1a1ow.png)
Therefore, the correct answer is 560 ways.