In an election, a voter may vote for any number of candidates, not greater than the number to be
elected. There are 10 candidates and 4 are to be elected. If a voter votes for at-least one
candidate, then the number of ways in which he can vote, is
(A) 385 (B) 1110 (C) 5040 (D) 6210

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Solution

Dear Student,

$A v o t e r c a n s e l e c t a n y c a n d i d a t e o u t o f 10 a n d h e h a s t o s e l e c t e a t l e a s t 1 c a n d i d a t e H e n c e t o t a l n u m b e r o f w a y s = o n l y 1 c a n d i d a t e + o n l y 2 c a n d i d a t e + o n l y 3 c a n d i d a t e + o n l y 4 c a n d i d a t e = {}^{10}C_{1} + {}^{10}C_{2} + {}^{10}C_{3} + {}^{10}C_{4} = 10 + \frac{10 \times 9}{2} + \frac{10 \times 9 \times 8}{6} + \frac{10 \times 9 \times 8 \times 7}{24} = 10 + 45 + 120 + 210 = 385$

Regards,

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In an election, a voter may vote for any number of candidates, not greater than the number to be elected. There are 10 candidates and 4 are to be elected. If a voter votes for at-least one candidate, then the number of ways in which he can vote, is (A) 385 (B) 1110 (C) 5040 (D) 6210

Dear Student, A voter can select any candidate out of 10 and he has to selecte atleast 1 candidateHence total number of ways=only 1 candidate+only 2 candidate+only 3 candidate+only 4 candidate=C110+C210+C310+C410=10+10×92+10×9×86+10×9×8×724=10+45+120+210=385 Regards,

In an election, a voter may vote for any number of candidates, not greater than the number to be
elected. There are 10 candidates and 4 are to be elected. If a voter votes for at-least one
candidate, then the number of ways in which he can vote, is
(A) 385 (B) 1110 (C) 5040 (D) 6210