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**Hint:**Here we have to find how many committees of four can be made with at least one lawyer and one chartered accountant in each committee. To do this we consider two cases, one where the person who is both lawyer and chartered accountant is included in the committee and another case where that person is not included. Using this we can find the number of committees possible as such.

**Complete step-by-step solution:**

We consider two cases here

Case(1): Where a person who is both lawyer and chartered accountant is included in the committee.

Then we have both a lawyer and a chartered accountant in the committee. We just have to select three persons of the remaining ten persons as,

\[

{}^{10}{C_3} = \dfrac{{10!}}{{3!7!}} \\

\Rightarrow {}^{10}{C_3} = \dfrac{{10 \times 9 \times 8 \times 7!}}{{3 \times 2 \times 7!}} \\

\Rightarrow {}^{10}{C_3} = \dfrac{{10 \times 9 \times 8}}{{3 \times 2}} \\

\Rightarrow {}^{10}{C_3} = 120 \]

So, \[120\] such possible committees

Case(2): When that person is not included

Selecting one lawyer and three chartered accountant\[ = {}^5{C_1} \times {}^5{C_3}\]

Selecting two lawyer and two chartered accountant \[ = {}^5{C_2} \times {}^5{C_2}\]

Selecting three lawyer and one chartered accountant\[ = {}^5{C_3} \times {}^5{C_1}\]

So total such committees are as,

\[= {}^5{C_1} \times {}^5{C_3} + {}^5{C_2} \times {}^5{C_2} + {}^5{C_3} \times {}^5{C_1} \\

= \dfrac{{5!}}{{4!1!}} \times \dfrac{{5!}}{{3!2!}} + \dfrac{{5!}}{{3!2!}} \times \dfrac{{5!}}{{3!2!}} + \dfrac{{5!}}{{3!2!}} \times \dfrac{{5!}}{{4!1!}} \\

= \dfrac{{5 \times 4!}}{{4!}} \times \dfrac{{5 \times 4 \times 3!}}{{2 \times 3!}} + \dfrac{{5 \times 4 \times 3!}}{{2 \times 3!}} \times \dfrac{{5 \times 4 \times 3!}}{{2 \times 3!}} + \dfrac{{5 \times 4 \times 3!}}{{2 \times 3!}} \times \dfrac{{5 \times 4!}}{{4!}} \\

= 5 \times 10 + 10 \times 10 + 10 \times 5 \\

= 200 \]

Then we add the result of both cases as,

\[200 + 120 = 320\]

**Hence we get that \[320\] such committees can be formed.**

**Note:**This is to note that the combination is used here to find the number of cases. Here \[{}^a{C_b}\] means that out of the total \[a\] things we have to select \[b\] things without any order. Hence this has helped us in cut shorting our calculations.

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