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In-how-many-ways-can-committee-of-5-be-formed-from-a-group-of-11-people-consisting-of-4-teachers-and-7-students-if-there-is-no-restriction-in-the-selection-




Question Number 142618 by iloveisrael last updated on 03/Jun/21
   In how many ways can committee  of 5 be formed from a group   of 11 people consisting of 4 teachers  and 7 students if there is no   restriction in the selection ?  _______________________
$$\:\:\:{In}\:{how}\:{many}\:{ways}\:{can}\:{committee} \\ $$$${of}\:\mathrm{5}\:{be}\:{formed}\:{from}\:{a}\:{group}\: \\ $$$${of}\:\mathrm{11}\:{people}\:{consisting}\:{of}\:\mathrm{4}\:{teachers} \\ $$$${and}\:\mathrm{7}\:{students}\:{if}\:{there}\:{is}\:{no}\: \\ $$$${restriction}\:{in}\:{the}\:{selection}\:? \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$
Commented by som(math1967) last updated on 03/Jun/21
C_5 ^(11) =((11×10×9×8×7×6!)/(5×4×3×2×6!))=22×3×7=462
$$\overset{\mathrm{11}} {{C}}_{\mathrm{5}} =\frac{\mathrm{11}×\mathrm{10}×\mathrm{9}×\mathrm{8}×\mathrm{7}×\mathrm{6}!}{\mathrm{5}×\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{6}!}=\mathrm{22}×\mathrm{3}×\mathrm{7}=\mathrm{462} \\ $$
Commented by otchereabdullai@gmail.com last updated on 04/Jun/21
nice
$$\mathrm{nice} \\ $$

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