I don’t understand what this is saying?

The binomial coefficients of the terms which are equidistant from the starting and the end are always equal. The simple reason behind this is

C(n, r) = C(n, n-r) which gives C(n, n) C(n, 1) = C(n, n-1) C(n, 2) = C(n, n-2).

Question

I don’t understand what this is saying?

The binomial coefficients of the terms which are equidistant from the starting and the end are always equal. The simple reason behind this is

C(n, r) = C(n, n-r) which gives C(n, n) C(n, 1) = C(n, n-1) C(n, 2) = C(n, n-2).

Nishant Vora · Answer

Means first binomial coefficient = last binomial coefficient 2nd binomial coefficient =2nd last binomial coefficient and so on Thanks

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I don’t understand what this is saying? The binomial coefficients of the terms which are equidistant from the starting and the end are always equal. The simple reason behind this is C(n, r) = C(n, n-r) which gives C(n, n) C(n, 1) = C(n, n-1) C(n, 2) = C(n, n-2).

I don’t understand what this is saying?

The binomial coefficients of the terms which are equidistant from the starting and the end are always equal. The simple reason behind this is

C(n, r) = C(n, n-r) which gives C(n, n) C(n, 1) = C(n, n-1) C(n, 2) = C(n, n-2).

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I don’t understand what this is saying? The binomial coefficients of the terms which are equidistant from the starting and the end are always equal. The simple reason behind this is C(n, r) = C(n, n-r) which gives C(n, n) C(n, 1) = C(n, n-1) C(n, 2) = C(n, n-2).

I don’t understand what this is saying? The binomial coefficients of the terms which are equidistant from the starting and the end are always equal. The simple reason behind this is C(n, r) = C(n, n-r) which gives C(n, n) C(n, 1) = C(n, n-1) C(n, 2) = C(n, n-2).

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I don’t understand what this is saying?

The binomial coefficients of the terms which are equidistant from the starting and the end are always equal. The simple reason behind this is

C(n, r) = C(n, n-r) which gives C(n, n) C(n, 1) = C(n, n-1) C(n, 2) = C(n, n-2).