As many as _________ different combination of water are possible from $_{1} H^{1},_{1} H^{2},_{1} H^{3}$ and $_{8} O^{16},_{8} O^{17}$ and $_{8} O^{18}$ .

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Solution

There are $n = 3$ different hydrogen isotopes to choose from, and we have to choose $r = 2$ of them.
Therefore,
$^{n + r - 1} C_{r} =^{2 + 3 - 1} C_{2} =^{4} C_{2} = 6$
Also we have $3$ different oxygen.
Therefore,
Total no. of different combinations of water $= 6 \times 3 = 18$

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As many as _________ different combination of water are possible from 1H1,1H2,1H3 and 8O16,8O17 and 8O18.

There are n=3 different hydrogen isotopes to choose from, and we have to choose r=2 of them.Therefore,n+r−1Cr=2+3−1C2=4C2=6Also we have 3 different oxygen.Therefore,Total no. of different combinations of water =6×3=18

As many as _________ different combination of water are possible from $_{1} H^{1},_{1} H^{2},_{1} H^{3}$ and $_{8} O^{16},_{8} O^{17}$ and $_{8} O^{18}$ .

There are $n = 3$ different hydrogen isotopes to choose from, and we have to choose $r = 2$ of them.
Therefore,
$^{n + r - 1} C_{r} =^{2 + 3 - 1} C_{2} =^{4} C_{2} = 6$
Also we have $3$ different oxygen.
Therefore,
Total no. of different combinations of water $= 6 \times 3 = 18$