Question

If A, B, and C are integers between 1 and 10 (inclusive), how many different combinations of A, B, and C exist such that A

Answer 1

$\fontsize{18}{10}{\textup{\textbf{The number of different combinations is 120.}}}$

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

$n_1=1+2+3+4+5+6+7+8=\dfrac{8(8+1)}{2}=36.$

Similarly, if A = 2, then the number of combinations is

$n_2=1+2+3+4+5+6+7=\dfrac{7(7+1)}{2}=28.$

If A = 3, then the number of combinations is

$n_3=1+2+3+4+5+6=\dfrac{6(6+1)}{2}=21.$

If A = 4, then the number of combinations is

$n_4=1+2+3+4+5=\dfrac{5(5+1)}{2}=15.$

If A = 5, then the number of combinations is

$n_5=1+2+3+4=\dfrac{4(4+1)}{2}=10.$

If A = 6, then the number of combinations is

$n_6=1+2+3=\dfrac{3(3+1)}{2}=6.$

If A = 7, then the number of combinations is

$n_7=1+2=\dfrac{2(2+1)}{2}=3.$

If A = 3, then the number of combinations is

$n_8=1.$

Therefore, the total number of combinations is

$n\\\\=n_1+n_2+n_3+n_4+n_5+n_6+n_7+n_8\\\\=36+28+21+15+10+6+3+1\\\\=120.$

Thus, the required number of different combinations is 120.

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