Evaluate
(i) `""^(61)C_(57)- ""^(60)C_(56)`
(ii) `""^(31)C_(26)- ""^(30)C_(26)`

Answer

To solve the given problems step by step, we will evaluate each part separately.

### Part (i): Evaluate \( \binom{61}{57} - \binom{60}{56} \)

1. **Use the identity**: We know that \( \binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1} \). Here, we can set \( n = 60 \) and \( r = 56 \).
\[ \binom{60}{56} + \binom{60}{57} = \binom{61}{57} \]

2. **Rearranging the equation**: From the above identity, we can express \( \binom{61}{57} \):
\[ \binom{61}{57} = \binom{60}{56} + \binom{60}{57} \]

3. **Substituting back into the original expression**:
\[ \binom{61}{57} - \binom{60}{56} = \binom{60}{57} \]

4. **Calculating \( \binom{60}{57} \)**:
\[ \binom{60}{57} = \binom{60}{3} = \frac{60!}{3!(60-3)!} = \frac{60!}{3! \cdot 57!} \]
\[ = \frac{60 \times 59 \times 58}{3 \times 2 \times 1} = \frac{60 \times 59 \times 58}{6} \]

5. **Calculating the product**:
\[ 60 \times 59 = 3540 \]
\[ 3540 \times 58 = 205320 \]
\[ \frac{205320}{6} = 34220 \]

Thus, the answer for part (i) is:
\[ \boxed{34220} \]

### Part (ii): Evaluate \( \binom{31}{26} - \binom{30}{26} \)

1. **Use the identity**: We cannot directly apply the previous identity since \( r \) is the same in both terms. Instead, we will calculate each term separately.

2. **Calculating \( \binom{31}{26} \)**:
\[ \binom{31}{26} = \binom{31}{5} = \frac{31!}{5!(31-5)!} = \frac{31!}{5! \cdot 26!} \]
\[ = \frac{31 \times 30 \times 29 \times 28 \times 27}{5 \times 4 \times 3 \times 2 \times 1} \]

3. **Calculating \( \binom{30}{26} \)**:
\[ \binom{30}{26} = \binom{30}{4} = \frac{30!}{4!(30-4)!} = \frac{30!}{4! \cdot 26!} \]
\[ = \frac{30 \times 29 \times 28 \times 27}{4 \times 3 \times 2 \times 1} \]

4. **Finding the common factor**:
Both terms share \( 26! \) in the denominator. Thus, we can factor it out:
\[ \binom{31}{26} - \binom{30}{26} = \frac{31 \times 30 \times 29 \times 28 \times 27}{5!} - \frac{30 \times 29 \times 28 \times 27}{4!} \]

5. **Factoring out common terms**:
\[ = \frac{30 \times 29 \times 28 \times 27}{4!} \left( \frac{31}{5} - 1 \right) \]
\[ = \frac{30 \times 29 \times 28 \times 27}{24} \left( \frac{31 - 5}{5} \right) = \frac{30 \times 29 \times 28 \times 27}{24} \cdot \frac{26}{5} \]

6. **Calculating the product**:
First calculate \( 30 \times 29 \times 28 \times 27 \):
\[ 30 \times 29 = 870 \]
\[ 870 \times 28 = 24360 \]
\[ 24360 \times 27 = 657720 \]
Now divide by \( 24 \):
\[ \frac{657720}{24} = 27405 \]
Finally multiply by \( \frac{26}{5} \):
\[ 27405 \times \frac{26}{5} = 142626 \]

Thus, the answer for part (ii) is:
\[ \boxed{142626} \]