Question

In how many ways can a party of 4 be selected out of 9, if one man is always to be
(i) included
(ii) excluded.​

Answer 1

SOLUTION

TO DETERMINE

The number of ways a party of 4 men can be selected out of 9 men , if one man is always to be

(i) included

(ii) excluded

EVALUATION

(i) Here total number of men = 9

Number of men to be selected = 4

Now this will be done such that one man is always to be included

Then ( 4 - 1 ) = 3 men are to be selected from

( 9 - 1 ) = 8 men

Hence the total number of selection

$\sf{ = {}^{8}C_3 }$

$\displaystyle \sf{ = \frac{8 \:! }{5! \: 3!} }$

$\displaystyle \sf{ = \frac{8 \times 7 \times 6 \times 5! }{5! \times \: 3 \times 2 \times 1} }$

$\displaystyle \sf{ = \frac{8 \times 7 \times 6 \times }{ \: 3 \times 2 \times 1} }$

$= 56$

(ii) Here total number of men = 9

Number of men to be selected = 4

Now this will be done such that one man is always to be excluded

Then 4 men are to be selected from ( 9 - 1 ) = 8 men

Hence the total number of selection

$\sf{ = {}^{8}C_4 }$

$\displaystyle \sf{ = \frac{8 \:! }{4! \: 4!} }$

$\displaystyle \sf{ = \frac{8 \times 7 \times 6 \times 5 \times 4! }{4! \times 4 \times \: 3 \times 2 \times 1} }$

$\displaystyle \sf{ = \frac{8 \times 7 \times 6 \times 5 }{ 4 \times \: 3 \times 2 \times 1} }$

$= 70$

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