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There are 10 men. 5 of them are to be selected for a team. In how many ways this can be done if a particular person is to be selected in the team and if A is selected then B also is to be selected?

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

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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 = $^7C_2+^8C_4$
$=21+70=91$
answered Sep 6, 2013 by rvidyagovindarajan_1
 

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