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Question Number 1164 by 112358 last updated on 08/Jul/15

$$Howmanyfivedigitnumbers$$$$existsuchthatthesumoftheir$$$$digitsequals43?$$$$Howmanyexistifthesumis$$$$39?$$$$$$

Commented by 123456 last updated on 08/Jul/15

$$a+b+c+d+e=k$$$$a\in \{1,2,3,...,9\}$$$$b,c,d,e\in \{0,1,...,9\}$$$$k\in \{39,43\}$$$${\mathrm{S}}_1\to k=39$$$${\mathrm{S}}_2\to k=43$$

Answered by prakash jain last updated on 09/Jul/15

$$\mathrm{Five}\mathrm{digits}\mathrm{all}9\mathrm{s}=45$$$$\mathrm{To}\mathrm{get}43\mathrm{we}\mathrm{need}\mathrm{to}\mathrm{subtract}2.$$$$00011\mathrm{can}\mathrm{be}\mathrm{subtracted}\mathrm{in}\frac{5!}{3!2!}\mathrm{ways}.$$$$00002\mathrm{can}\mathrm{be}\mathrm{subtracted}\mathrm{in}\frac{5!}{4!1!}\mathrm{ways}.$$$$\mathrm{Total}\mathrm{ways}\mathrm{to}\mathrm{form}5\mathrm{digit}\mathrm{number}\mathrm{with}$$$$\mathrm{sum}\mathrm{of}\mathrm{digits}43=\frac{5!}{3!2!}+\frac{5!}{4!1!}=10+5=15$$

Commented by prakash jain last updated on 09/Jul/15

$$88999898998998989998$$$$988999898998998$$$$9988999898$$$$99988$$$$7999997999997999997999997$$

Answered by prakash jain last updated on 09/Jul/15

$$\mathrm{To}\mathrm{get}\mathrm{sum}39\mathrm{we}\mathrm{need}\mathrm{to}\mathrm{subtract}6.$$$$6\mathrm{can}\mathrm{be}\mathrm{obtained}\mathrm{by}$$$$41\mathrm{s}\mathrm{and}12\left(11112\right)=\frac{5!}{4!1!}=5$$$$31\mathrm{s},10\mathrm{and}13\left(11103\right)=\frac{5!}{3!1!1!}=20$$$$21\mathrm{s},10\mathrm{and}22\mathrm{s}\left(11022\right)=\frac{5!}{2!2!1!}=30$$$$21\mathrm{s},20\mathrm{s}\mathrm{and}14\left(11004\right)=\frac{5!}{2!2!1!}=30$$$$11\mathrm{s},20\mathrm{s}\mathrm{and}12,13\left(10023\right)=\frac{5!}{2!1!1!1!}=60$$$$11\mathrm{s},30\mathrm{s}\mathrm{and}15\left(10005\right)=\frac{5!}{1!3!1!}=20$$$$01\mathrm{s},32\mathrm{s}\mathrm{and}20\mathrm{s}\left(22200\right)=\frac{5!}{2!3!}=10$$$$01\mathrm{s},12,30\mathrm{and}14\left(20004\right)=\frac{5!}{3!}=20$$$$01\mathrm{s},02\mathrm{s},23\mathrm{s}\left(00033\right)=\frac{5!}{3!2!}=10$$$$01\mathrm{s},02\mathrm{s},16\left(00006\right)=\frac{5!}{1!4!}=5$$$$\mathrm{Total}=\mathrm{sum}\mathrm{of}\mathrm{all}\mathrm{above}.$$$$$$

Commented by 112358 last updated on 09/Jul/15

$$Thanks$$

Commented by 112358 last updated on 09/Jul/15

$$Thereareacouplecomputation$$$$errorsbutIunderstandthe$$$$solution.$$

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