Total number of selecting three partners randomly in which at least one junior partner must be there out of 23 senior and 24 junior partners is 14,444
Combination:
Combination is the number of possible ways, where order is irrelevant and replacements are not permitted, to select a sample of r elements from a set of n different objects.
[tex]C(n,r)=nCr = \frac{n!}{(n-r)! r!}[/tex]
C(n, r) means, Selection of r things out of n different things
According to question,
Given,
A firm has 23 senior and 24 junior partners.
We have to select three partners randomly in which at least one junior partner must be there.
Case I : 2 senior and 1 junior partners
2 senior partners can be selected out of 23 in C(23,2) ways
1 junior partner can be selected out of 24 in C(24,1) ways
hence number of ways of selecting 2 senior and 1 junior partners is equal to
C(23,2) x C(24,1)
Case II : 2 senior and 1 junior partners
1 senior partners can be selected out of 23 in C(23,1) ways
2 junior partner can be selected out of 24 in C(24,2) ways
hence number of ways of selecting 1 senior and 2 junior partners is equal to
C(23,1) x C(24,2)
Case III : 0 senior and 3 junior partners
0 senior partners can be selected out of 23 in C(23,0) ways
3 junior partner can be selected out of 24 in C(24,3) ways
hence number of ways of selecting no senior and 3 junior partners is equal to
C(23,0) x C(24,3)
Thus,
Total number of selecting three partners randomly in which at least one junior partner must be there is
C(23,2) x C(24,1) + C(23,1) x C(24,2) + C(23,0) x C(24,3)
Using formula,
[tex]C(n,r)=nCr = \frac{n!}{(n-r)! r!}[/tex]
C(23,2) x C(24,1) + C(23,1) x C(24,2) + C(23,0) x C(24,3)
= 253 x 24 + 23 x 276 + 1 x 2024
= 6072 + 6348 + 2024
= 14,444
Total number of selecting three partners randomly in which at least one junior partner must be there is 14,444
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