The number of ways to distribute 30 identical candies among four children C_1,C_2,C_3 and C_4 so that C_2 receives atleast 4 and atmost 7 candies,…

Given, ${C}_{1}+{C}_{2}+{C}_{3}+{C}_{4}=30$C1​+C2​+C3​+C4​=30

Now it has given that 4\leq {C}_{2}\leq 7&2\leq {C}_{3}\leq 6

and {C}_{1}&{C}_{4} can take any value.

Finding coefficient of ${x}^{30}$x30 in

$$({x}^{0}+{x}^{1}+......{x}^{30})({x}^{4}+{x}^{5}+......{x}^{7})({x}^{2}+...{x}^{6})({x}^{0}+....{x}^{30})$$(x0+x1+......x30)(x4+x5+......x7)(x2+...x6)(x0+....x30)

$$={(\frac{1-{x}^{31}}{1-x})}^{2}{x}^{4}(\frac{1-{x}^{4}}{1-x}){x}^{2}(\frac{1-{x}^{5}}{1-x})$$=(1−x1−x31​)2x4(1−x1−x4​)x2(1−x1−x5​)

{Ignoring higher power more than $30$30}

$$={(\frac{1}{1-x})}^{2}{x}^{4}(\frac{1-{x}^{4}}{1-x}){x}^{2}(\frac{1-{x}^{5}}{1-x})$$=(1−x1​)2x4(1−x1−x4​)x2(1−x1−x5​)

$={x}^{6}(1-{x}^{4})(1-{x}^{5}){(1-x)}^{-4}$=x6(1−x4)(1−x5)(1−x)−4 $={x}^{6}(1-{x}^{5}-{x}^{4}+{x}^{9}){(1-x)}^{-4}$=x6(1−x5−x4+x9)(1−x)−4

$$=({x}^{6}-{x}^{11}-{x}^{10}+{x}^{15}){(1-x)}^{-4}$$=(x6−x11−x10+x15)(1−x)−4

Required coefficient $=C244+24-1-C194+19-1$=C244+24−1−C194+19−1 $-C2020+4-1+C1515+4-1$−C2020+4−1+C1515+4−1

$=C2427-C1922-C2023+C1518$=C2427−C1922−C2023+C1518 $=430$=430