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Each pair has to be of a different kind, and the seventh card must also be a different kind.

Here is my attempt: Choose 1 card from the 52, then take 1 of 3 possible matches, then choose another from the remaining 48, then do the same thing, and then repeat the process with 44, and finally choose 1 of the remaining 40 cards. This yields:

$52$ x $3$ x $48$ x $3$ x $44$ x $3$ x $40$ = $118,609,920$ possible hands

Is this correct?

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    That number is far too high. There are only $\binom {52}7= 133,784,560$ possible seven card hands. – 2017-01-24
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    You are multiply counting...for example, say we first choose the $A\spadesuit$ and then $K\spadesuit$ to make the pair. That's the same as if we made the choice in the opposite order. – 2017-01-24
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    First choose 3 ranks from 13, then choose 2 out of 4 cards of each of the three chosen ranks, and finally choose one card from the remaining 40 cards. – 2017-01-24

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First pick the three denominations to be paired. There are ${13\choose 3}$ ways to do this. Then, for each denomination, pick a pair. There are ${4\choose 2}$ ways to do this, for each of the three denominations. Lastly, there are $52-12=40$ cards in the other 10 denominations. We need one of them for the seventh card. Putting it all together we get $${13\choose 3}{4\choose 2}^340=2471040$$