$\begin {array} {1 1} (A)\;21 & \quad (B)\;56 \\ (C)\;91 & \quad (D)\;126 \end {array}$

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Out of 5 to be selected from 10, one particular man is to be selected.

$\therefore$ 4 are to be selected from 9.

Case(1) A is selected. $\therefore$ B is also selected.

$\therefore$ 3 out of 5 are selected already. (A,B, one particular man)

Any two are to be selected from 7 in $^7C_2$ ways.

Case (ii) If A is not selected

then B may or may not be selected.

Remaining 4 are to be selected from 8 (excluding A)

This can be done in $^8C_4$ ways.

$\therefore$ Required no. of selection = $^7C_2+^8C_4$

$=21+70=91$

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