Problem 4. In a box, there are 50 cards on which the first 100 non-zero natural numbers are written, as follows: on the first card, the numbers 1 (on one side) and 2 (on the other side) are written, on the second card, the numbers 3 (on one side) and 4 (on the other side) are written, and so on, up to the 50th card, on which the numbers 99 (on one side) and 100 (on the other side) are written.

Eliza takes out four cards from the box and calculates the sum of the eight numbers written on them. How many distinct sums can Eliza obtain?
Solution. The sums of the two numbers on each card are: $3,7,11,15, \ldots, 195,199$, i.e., numbers of the form $4 k-1$, where $k$ is a natural number taking values between 1 and $50 \ldots \ldots \ldots .2$ p

If $4 a-1,4 b-1,4 c-1$ and $4 d-1$ are the sums on the four cards, the total sum is $4(a+b+c+d)-4$, so the problem reduces to calculating the number of values that the sum of four natural numbers $a<b<c<d$, between 1 and 50, can take.

The minimum sum is $1+2+3+4=10$, and the maximum sum is $47+48+49+50=194 \ldots \ldots \mathbf{1}$ p

We will show that each sum $a+b+c+d$ can take all values from 10 to 194, i.e., a total of 185 values. Thus we have:

- sums from 10 to 56 are obtained for $a=1, b=2, c=3$ and $d$ taking values from 4 to 50: $1+2+3+4,1+2+3+5,1+2+3+6, \ldots, 1+2+3+50$
- sums from 57 to 102 are obtained for $a=1, b=2, d=50$ and $c$ taking values from 4 to 49: $1+2+4+50,1+2+5+50,1+2+6+50, \ldots, 1+2+49+50$
- sums from 103 to 148 are obtained for $a=1, c=49, d=50$ and $b$ taking values from 3 to 48: $1+3+49+50,1+4+49+50,1+5+49+50, \ldots, 1+48+49+50$
- sums from 149 to 194 are obtained for $b=48, c=49, d=50$ and $a$ taking values from 2 to 47: $2+48+49+50,3+48+49+50,4+48+49+50, \ldots, 47+48+49+50$

In conclusion, Eliza can obtain 185 different sums, namely all numbers between

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