Respuesta :
[tex]\fontsize{18}{10}{\textup{\textbf{The number of different combinations is 120.}}}[/tex]
Step-by-step explanation:
A, B and C are integers between 1 and 10 such that A<B<C.
The value of A can be minimum 1 and maximum 8.
If A = 1, B = 2, then C can be one of 3, 4, 5, 6, 7, 8, 9, 10 (8 options).
If A = 1, B = 3, then C has 7 options (4, 5, 6, 7, 8, 9, 10).
If A = 1, B = 4, then C has 6 options (5, 6, 7, 8, 9, 10).
If A = 1, B = 5, then C has 5 options (6, 7, 8, 10).
If A = 1, B = 6, then C has 4 options (7, 8, 9, 10).
If A = 1, B = 7, then C has 3 options (8, 9, 10).
If A = 1, B = 8, then C has 2 options (9, 10).
If A = 1, B = 9, then C has 1 option (10).
So, if A = 1, then the number of combinations is
[tex]n_1=1+2+3+4+5+6+7+8=\dfrac{8(8+1)}{2}=36.[/tex]
Similarly, if A = 2, then the number of combinations is
[tex]n_2=1+2+3+4+5+6+7=\dfrac{7(7+1)}{2}=28.[/tex]
If A = 3, then the number of combinations is
[tex]n_3=1+2+3+4+5+6=\dfrac{6(6+1)}{2}=21.[/tex]
If A = 4, then the number of combinations is
[tex]n_4=1+2+3+4+5=\dfrac{5(5+1)}{2}=15.[/tex]
If A = 5, then the number of combinations is
[tex]n_5=1+2+3+4=\dfrac{4(4+1)}{2}=10.[/tex]
If A = 6, then the number of combinations is
[tex]n_6=1+2+3=\dfrac{3(3+1)}{2}=6.[/tex]
If A = 7, then the number of combinations is
[tex]n_7=1+2=\dfrac{2(2+1)}{2}=3.[/tex]
If A = 3, then the number of combinations is
[tex]n_8=1.[/tex]
Therefore, the total number of combinations is
[tex]n\\\\=n_1+n_2+n_3+n_4+n_5+n_6+n_7+n_8\\\\=36+28+21+15+10+6+3+1\\\\=120.[/tex]
Thus, the required number of different combinations is 120.
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Question : ow many types of zygotic combinations are possible between a cross AaBBCcDd × AAbbCcDD?
Link : https://brainly.in/question/4909567.
