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Question

Let $$5 < n_1 < n_2 < n_3 < n_4$$ be integers such that $$n_1+n_2+n_3+n_4=35$$. The number of such distinct arrangements $$(n_1, n_2, n_3, n_4)$$.


A
38C3
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B
8C3
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C
5
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D
6
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Solution

The correct option is A $$^{38}C_3$$
$${ n }_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 }+.....+{ n }_{ k }=R$$
for this arrangement,
No. of arrangement or distinct arrangements are $${ R+K-1 }_{ { C }_{ K-1 } }$$
So, for $${ n }_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 }+{ n }_{ 4 }=35$$
No. of arrangements $$={ 35+4-1 }_{ { C }_{ 4-1 } }={ 38 }_{ { C }_{ 3 } }$$

Mathematics

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