The number of ways to distribute 30 identical candies among four children C1, C2, C3 and C4 so that C2 receives atleast 4 and atmost 7 candies, C3 receives atleast 2 and atmost 6 candies, is equal to:
The correct answer is (D) : 430
By multinomial theorem, no. of ways to distribute 30 identical candies among four children C1, C2 and C3, C4
= Coefficient of x30 in (x4 + x5 + … + x7) (x2 + x3 +…+ x6) (1 + x + x2…)2
=Coefficient of x24 in \(\frac{(1−x^4)(1−x^5)(1−x^{31})^2}{(1−x)(1-x)(1-x)^2}\)
= Coefficient of x24 in (1 – x4 – x5 + x9) (1 – x)–4
= 27C24 – 23C20 – 22C19 + 18C15 = 430
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Then the number of elements $ x $ in $ S $ such that at least one of the digits 0 and 1 appears exactly twice in $ x $, is equal to __________.
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In the given circuit the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 Ω, the value of the current in the circuit will be A.
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