Answer: b) 100
Step-by-step explanation:
Given : Number of senior partners =4
Number of junior partners. = 6
Total partners = 4+ 6= 10
Total Number of ways to choose 3 partners[tex]=^{10}C_3=\dfrac{10!}{3!(10-3)!}[/tex]
[tex][\because\ ^nC_r=\dfrac{n!}{r!(n-r)!}][/tex]
[tex]=\dfrac{10\times9\times8\times7!}{7!\times6}=120[/tex]
i.e. Total Number of ways to choose 3 partners =120
The number of ways that none of 3 partners are seniors =tex]=^{6}C_3=\dfrac{6!}{3!(6-3)!}=20[/tex]
Now , the different groups of 3 partners can be formed in which at least one member of the group is a senior partner
= Total Number of ways to choose 3 partners- Number of ways that none of 3 partners are seniors
= 120-20=100
Hence, the correct answer is b) 100.
