From $1$ to $50$, take a pair of integers (repetition allowed) so that their sum is greater than $50$ , how many ways are there pick such pair?
Attempt:
1st number = $50$ , 2nd number ($1$ to $50$) - $50$ combinations
1st number = $49$, 2nd number ($2$ to $49$) - $49$ combinations
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1st number number = $1$ , 2nd number $50$ - $1$ combination
Summation of $1$ to $50 = 1275$
On
Your solution is correct, but only if you have to verify this fact: Repetition means choosing the same two numbers is allowed, does it mean that you can choose the same pair, swapped positions as well? (For example, $50$ and $1$, $1$ and 50). If the same pair, swapped positions is not allowed then the answer must be reduced:
Assume that the first number is always less than or equal the second number.
1st number = $1$; 2nd number = $50$ - $1$ combination
1st number = $2$; 2nd number ($49$ to $50$) - $2$ combinations
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1st number = $24$; 2nd number ($27$ to $50$) - $24$ combinations
1st number = $25$; 2nd number ($26$ to $50$) - $25$ combinations
1st number = $26$; 2nd number ($26$ to $50$) - $25$ combinations
1st number = $27$; 2nd number ($27$ to $50$) - $24$ combinations
1st number = $28$; 2nd number ($28$ to $50$) - $23$ combinations
1st number = $29$; 2nd number ($29$ to $50$) - $22$ combinations
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1st number = $49$; 2nd number ($49$ to $50$) - $2$ combinations
1st number = $50$; 2nd number = $50$ - $1$ combination
If the same pair, swapped positions is not allowed:
$(1+2+3+...+50)\times 2=650$ combinations
When they say 'pair', I am sure the order does not matter. So, you are double-counting many pairs. How many? Well, pretty much all of them, except the ones where the two numbers are the same.
Since you have $25$ pairs with the same number - namely $(26,26)$ through $(50,50)$ - that means you have $1275-25=1250$ ordered pairs with two different numbers, and hence only $\frac{1250}{2}=625$ unordered pairs.
Total: $625+25=650$ unordered pairs